RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions

RD Sharma Solutions Class 10 Maths Chapter 9 – Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 9 – Arithmetic Progressions are provided here for students to study and excel in their board exams. In order to succeed in the Class 10 Mathematics examinations, a structured way of understanding the concepts and solving problems is a must. The RD Sharma Solutions are on par with these needs for a student to secure good marks in the qualification exam. Furthermore, the solutions have been created by our expert team at BYJU’S, which are in simple language with appropriate explanations to give clarity on the concepts in the chapter.

Arithmetic Progressions of RD Sharma Solutions Class 10 is an exciting chapter where students can score good marks easily. But due to some misconceptions, not all students can grab those marks. In order to help students, the RD Sharma Solutions for Class 10 provides descriptive answers to all six exercises. This chapter’s key aspects are

• Understanding sequences
• Arithmetic Progressions
• Describing the sequence by writing the algebraic formula for its terms
• Find the sum of terms in an A.P.
• Solving various word problems related to Arithmetic Progressions

RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions

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Access the RD Sharma Solutions for Class 10 Maths Chapter 9 – Arithmetic Progression

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.1 Page No: 9.5

1. Write the first terms of each of the following sequences whose n^th term is

(i) a_n = 3n + 2

(ii) a_n = (n – 2)/3

(iii) a_n = 3ⁿ

(iv) a_n = (3n – 2)/ 5

(v) a_n = (-1)ⁿ . 2ⁿ

(vi) a_n = n(n – 2)/2

(vii) a_n = n² – n + 1

(viii) a_n = n² – n + 1

(ix) a_n = (2n – 3)/ 6

Solutions:

(i) a_n = 3n + 2

Given sequence whose a_n = 3n + 2

To get the first five terms of a given sequence, put n = 1, 2, 3, 4, 5, and we get

a₁ = (3 × 1) + 2 = 3 + 2 = 5

a₂ = (3 × 2) + 2 = 6 + 2 = 8

a₃ = (3 × 3) + 2 = 9 + 2 = 11

a₄ = (3 × 4) + 2 = 12 + 2 = 14

a₅ = (3 × 5) + 2 = 15 + 2 = 17

∴ the required first five terms of the sequence whose n^th term, a_n = 3n + 2 are 5, 8, 11, 14, 17.

(ii) a_n = (n – 2)/3

Given sequence whose

On putting n = 1, 2, 3, 4, 5, then can get the first five terms.

∴ the required first five terms of the sequence whose n^th term.

(iii) a_n = 3ⁿ

Given the sequence whose a_n = 3ⁿ

To get the first five terms of the given sequence, put n = 1, 2, 3, 4, 5 in the above

a₁ = 3¹ = 3;

a₂ = 3² = 9;

a₃ = 27;

a₄ = 3⁴ = 81;

a₅ = 3⁵ = 243.

∴ the required first five terms of the sequence whose n^th term, a_n = 3ⁿ are 3, 9, 27, 81, 243.

(iv) a_n = (3n – 2)/ 5

Given sequence whose

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

And we get

∴ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5

(v) a_n = (-1)ⁿ2ⁿ

Given sequence whose a_n = (-1)ⁿ2ⁿ

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

a₁ = (-1)¹.2¹ = (-1).2 = -2

a₂ = (-1)².2² = (-1).4 = 4

a₃ = (-1)³.2³ = (-1).8 = -8

a₄ = (-1)⁴.2⁴ = (-1).16 = 16

a₅ = (-1)⁵.2⁵ = (-1).32 = -32

∴ the first five terms of the sequence are – 2, 4, – 8, 16, – 32.

(vi) a_n = n(n – 2)/2

The given sequence is

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And we get

∴ the required first five terms are -1/2, 0, 3/2, 4, 15/2

(vii) a_n = n² – n + 1

The given sequence whose, a_n = n² – n + 1

To get the first five terms of the given sequence, put n = 1, 2, 3, 4, 5.

And we get

a₁ = 1² – 1 + 1 = 1

a₂ = 2² – 2 + 1 = 3

a₃ = 3² – 3 + 1 = 7

a₄ = 4² – 4 + 1 = 13

a₅ = 5² – 5 + 1 = 21

∴ the required first five terms of the sequence are 1, 3, 7, 13, 21.

(viii) a_n = 2n² – 3n + 1

The given sequence whose a_n = 2n² – 3n + 1

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And we get

a₁ = 2.1² – 3.1 + 1 = 2 – 3 + 1 = 0

a₂ = 2.2² – 3.2 + 1 = 8 – 6 + 1 = 3

a₃ = 2.3² – 3.3 + 1 = 18 – 9 + 1 = 10

a₄ = 2.4² – 3.4 + 1 = 32 – 12 + 1 = 21

a₅ = 2.5² – 3.5 + 1 = 50 – 15 + 1 = 36

∴ the required first five terms of the sequence are 0, 3, 10, 21, and 36.

(ix) a_n = (2n – 3)/ 6

Given the sequence whose,

To get the first five terms of the sequence, we put n = 1, 2, 3, 4, 5.

And we get

∴ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.2 Page No: 9.8

1. Show that the sequence defined by a_n = 5n – 7 is an A.P., find its common difference.

Solution:

Given, a_n = 5n – 7

Now, putting n = 1, 2, 3, 4, we get,

a₁ = 5(1) – 7 = 5 – 7 = -2

a₂ = 5(2) – 7 = 10 – 7 = 3

a₃ = 5(3) – 7 = 15 – 7 = 8

a₄ = 5(4) – 7 = 20 – 7 = 13

We can see that,

a₂ – a₁ = 3 – (-2) = 5

a₃ – a₂ = 8 – (3) = 5

a₄ – a₃ = 13 – (8) = 5

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = 5n – 7 is an A.P. with the common difference 5.

2. Show that the sequence defined by a_n = 3n² – 5 is not an A.P.

Solution:

Given, a_n = 3n² – 5

Now, putting n = 1, 2, 3, 4, we get,

a₁ = 3(1)² – 5= 3 – 5 = -2

a₂ = 3(2)² – 5 = 12 – 5 = 7

a₃ = 3(3)² – 5 = 27 – 5 = 22

a₄ = 3(4)² – 5 = 48 – 5 = 43

We can see that,

a₂ – a₁ = 7 – (-2) = 9

a₃ – a₂ = 22 – 7 = 15

a₄ – a₃ = 43 – 22 = 21

Since the difference between the terms is not common and varies, we can conclude that the given sequence defined by a_n = 3n² – 5 is not an A.P.

3. The general term of a sequence is given by a_n = -4n + 15. Is the sequence an A.P.? If so, find its 15^th term and the common difference.

Solution:

Given, a_n = -4n + 15

Now, putting n = 1, 2, 3, 4, we get,

a₁ = -4(1) + 15 = -4 + 15 = 11

a₂ = -4(2) + 15 = -8 + 15 = 7

a₃ = -4(3) + 15 = -12 + 15 = 3

a₄ = -4(4) + 15 = -16 + 15 = -1

We can see that,

a₂ – a₁ = 7 – (11) = -4

a₃ – a₂ = 3 – 7 = -4

a₄ – a₃ = -1 – 3 = -4

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = -4n + 15 is an A.P with a common difference of -4.

Hence, the 15^th term will be

a₁₅ = -4(15) + 15 = -60 + 15 = -45

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.3 Page No: 9.11

1. For the following arithmetic progressions, write the first term a and the common difference d.

(i) – 5, -1, 3, 7,…

(ii) 1/5, 3/5, 5/5, 7/5,…

(iii) 0.3, 0.55, 0.80, 1.05,…

(iv) -1.1, – 3.1, – 5.1, – 7.1,…

Solution:

We know that if a is the first term and d is a common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d,….

(i) – 5, –1, 3, 7,…

Given arithmetic series is – 5, –1, 3, 7,…

c a, a + d, a + 2d + a + 3d,….

Thus, by comparing these two, we get, a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7

First term (a) = – 5

By subtracting the second and first terms, we get

(a + d) – (a) = d

-1 – (- 5) = d

4 = d

⇒ Common difference (d) = 4.

(ii) 1/5, 3/5, 5/5, 7/5, ………….

Given arithmetic series is 1/5, 3/5, 5/5, 7/5, ……………

It is seen that it’s of the form of 1/5, 2/5, 5/5, 7/5, ……….. a, a + d, a + 2d, a + 3d,

Thus, by comparing these two, we get

a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5

First term (a) = 1/5

By subtracting the first term from the second term, we get

d = (a + d)-(a)

d = 3/5 – 1/5

d = 2/5

⇒ common difference (d) = 2/5

(iii) 0.3, 0.55, 0.80, 1.05, …………

Given arithmetic series 0.3, 0.55, 0.80, 1.05, ……….

It is seen that it’s in the form of a, a + d, a + 2d, a + 3d,

Thus, by comparing, we get,

a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05

First term (a) = 0.3.

By subtracting the first term from a second term, we get

d = (a + d) – (a)

d = 0.55 – 0.3

d = 0.25

⇒ Common difference (d) = 0.25

(iv) –1.1, – 3.1, – 5.1, –7.1, ……..

General series is –1.1, – 3.1, – 5.1, –7.1, ……..

It is seen that it’s of the form of a, a + d, a + 2d, a + 3d, ………..

Thus, by comparing these two, we get

a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –7.1

First term (a) = –1.1

Common difference (d) = (a + d) – (a)

= -3.1 – ( – 1.1)

⇒ Common difference (d) = – 2

2. Write the arithmetic progression when the first term a and common difference d are as follows:

(i) a = 4, d = – 3

(ii) a = –1, d = 1/2

(iii) a = –1.5, d = – 0.5

Solution:

We know that, if first term (a) = a and common difference = d, then the arithmetic series is: a, a + d, a + 2d, a + 3d,

(i) a = 4, d = -3

Given, the first term (a) = 4

Common difference (d) = -3

Then arithmetic progression is a, a + d, a + 2d, a + 3d, ……

⇒ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……

⇒ 4, 1, – 2, – 5, – 8 ……..

(ii) a = -1, d = 1/2

Given, the first term (a) = -1

Common difference (d) = 1/2

Then arithmetic progression is a, a + d, a + 2d, a + 3d,

⇒ -1, -1 + 1/2, -1 + 2½, -1 + 3½, …

⇒ -1, -1/2, 0, 1/2

(iii) a = –1.5, d = – 0.5

Given the first term (a) = –1.5

Common difference (d) = – 0.5

Then arithmetic progression is; a, a + d, a + 2d, a + 3d, ……

⇒ -1.5, -1.5 + (-0.5), –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)

⇒ – 1.5, – 2, – 2.5, – 3, …….

3. In which of the following situations the sequence of numbers formed will form an A.P.?
(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder. 

(iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of the first year, second year, third year, …, and so on.

Solution:

(i) Given,

Cost of digging a well for the first meter (c₁) = Rs.150.

And the cost rises by Rs.20 for each succeeding meter

Then,

Cost of digging for the second meter (c₂) = Rs.150 + Rs 20 = Rs 170

Cost of digging for the third meter (c₃) = Rs.170 + Rs 20 = Rs 210

Hence, its clearly seen that the costs of digging a well for different lengths are 150, 170, 190, 210, ….

Evidently, this series is in A∙P.

With first term (a) = 150, common difference (d) = 20

(ii) Given,

Let the initial volume of air in a cylinder be V litres each time 3^th/4 of air in a remaining i.e

1 -1/4

The first time, the air in the cylinder is V.

The second time, the air in the cylinder is 3/4 V.

The third time, the air in the cylinder is (3/4)² V.

Thus, series is V, 3/4 V, (3/4)² V,(3/4)³ V, ….

Hence, the above series is not an A.P.

(iii) Given,

Divya deposited Rs 1000 at compound interest of 10% p.a

So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100

And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210

And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331

Clearly, these amounts 1100, 1210 and 1331 are not in an A.P. since the difference between them is not the same.

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.4 Page No: 9.24

1. Find:

(i) 10^th tent of the AP 1, 4, 7, 10….

(ii) 18^th term of the AP √2, 3√2, 5√2, ……. 

(iii) n^th term of the AP 13, 8, 3, -2, ………. 

(iv) 10^th term of the AP -40, -15, 10, 35, …………. 

(v) 8^th term of the AP 11, 104, 91, 78, …………… 

(vi) 11^th tenor of the AP 10.0, 10.5, 11.0, 11.2, ………….. 

(vii) 9^th term of the AP 3/4, 5/4, 7/4 + 9/4, ………..

Solution:

(i) Given A.P. is 1, 4, 7, 10, ……….

First term (a) = 1

Common difference (d) = Second term – First term

= 4 – 1 = 3.

We know that, n^th term in an A.P = a + (n – 1)d

Then, 10^th term in the A.P is 1 + (10 – 1)3

= 1 + 9×3

= 1 + 27

= 28

∴ 10^th term of A. P. is 28

(ii) Given A.P. is √2, 3√2, 5√2, …….

First term (a) = √2

Common difference = Second term – First term

= 3√2 – √2

⇒ d = 2√2

We know that the n^th term in an A. P. = a + (n – 1)d

Then, the 18^th term of A. P. = √2 + (18 – 1)2√2

= √2 + 17.2√2

= √2 (1+34)

= 35√2

∴ 18^th term of A. P. is 35√2

(iii) Given A. P. is 13, 8, 3, – 2,  …………

First term (a) = 13

Common difference (d) = Second term first term

= 8 – 13 = – 5

We know that, the n^th term of an A.P. a_n = a +(n – 1)d

= 13 + (n – 1) – 5

= 13 – 5n + 5

∴ the n^th term of the A.P is a_n = 18 – 5n

(iv) Given A. P. is – 40, -15, 10, 35, ……….

First term (a) = -40

Common difference (d) = Second term – fast term

= -15 – (- 40)

= 40 – 15

= 25

We know that, the n^th term of an A.P. a_n = a + (n – 1)d

Then, the 10^th term of A. P. a₁₀ = -40 + (10 – 1)25

= – 40 + 9.25

= – 40 + 225

= 185

∴ the 10^th term of the A. P. is 185

(v) Given sequence is 117, 104, 91, 78, ………….

First term (a) = 117

Common difference (d) = Second term – first term

= 104 – 117

= – 13

We know that, the n^th term = a + (n – 1)d

Then, the 8^th term = a + (8 – 1)d

= 117 + 7(-13)

= 117 – 91

= 26

∴ the 8^th term of the A. P. is 26

(vi) Given A. P is 10.0, 10.5, 11.0, 11.5,

First term (a) = 10.0

Common difference (d) = Second term – first term

= 10.5 – 10.0 = 0.5

We know that, the n^th term a_n = a + (n – 1)d

Then, the 11^th term a₁₁ = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5

= 10.0 + 5

=15.0

∴ the 11^th term of the A. P. is 15.0

(vii) Given A. P is 3/4,  5/4, 7/4, 9/4, …………

First term (a) = 3/4

Common difference (d) = Second term – first term

= 5/4 – 3/4

= 2/4

We know that, the n^th term a_n = a + (n – 1)d

Then, the 9^th term a₉ = a + (9 – 1)d

∴ the 9^th term of the A. P. is 19/4.

2.(i) Which term of the AP 3, 8, 13, …. is 248?

(ii) Which term of the AP 84, 80, 76, … is 0?

(iii) Which term of the AP 4. 9, 14, …. is 254?

(iv) Which term of the AP 21. 42, 63, 84, … is 420?

(v) Which term of the AP 121, 117. 113, … is its first negative term?

Solution:

(i) Given A.P. is 3, 8, 13, ………..

First term (a) = 3

Common difference (d) = Second term – first term

= 8 – 3

= 5

We know that, the n^th term (a_n) = a + (n – 1)d

And, given the n^th term a_n = 248

248 = 3+(n – 1)5

248 = -2 + 5n

5n = 250

n =250/5 = 50

∴ the 50^th term in the A.P. is 248.

(ii) Given A. P is 84, 80, 76, …………

First term (a) = 84

Common difference (d) = a₂ – a

= 80 – 84

= – 4

We know that, the n^th term (a_n) = a +(n – 1)d

And, given the nth term is 0

0 = 84 + (n – 1) – 4

84 = +4(n –  1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

∴ the 22^nd term in the A.P is 0.

(iii) Given A. P 4, 9, 14, …………

First term (a) = 4

Common difference (d) = a₂ – a₁

= 9 – 4

= 5

We know that, the n^th term (a_n) = a + (n – 1)d

And, given the n^th term is 254

4 + (n – 1)5 = 254

(n – 1)∙5 = 250

n – 1 = 250/5 = 50

n = 51

∴  the 51st term in the A.P. is 254.

(iv) Given A. P 21, 42, 63, 84, ………

a = 21, d = a₂ – a₁

= 42 – 21

= 21

We know that, the n^th term (a_n) = a +(n – 1)d

And, given the n^th term = 420

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

∴ The 20^th term is 420.

(v) Given A.P is 121, 117, 113, ………..

Fiat term (a) = 121

Common difference (d) = 117 – 121

= – 4

We know that, the n^th term a_n = a + (n – 1)d

And, for some the n^th term is negative, i.e., a_n < 0

121 + (n – 1) – 4 < 0

121 + 4 – 4n < 0

125 – 4n < 0

4n > 125

n > 125/4

n > 31.25

The integer which comes after 31.25 is 32.

∴ The 32^nd term in the A.P. will be the first negative term.

3.(i) Is 68 a term of the A.P. 7, 10, 13,… ?

(ii) Is 302 a term of the A.P. 3, 8, 13, …. ?

(iii) Is -150 a term of the A.P. 11, 8, 5, 2, … ?

Solutions:

(i) Given, A.P. 7, 10, 13,…

Here, a = 7 and d = a₂ – a₁ = 10 – 7 = 3

We know that, the n^th term a_n = a + (n – 1)d

Required to check the n^th term a_n = 68

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n – 3 = 68

3n + 4 = 68

3n = 64

⇒ n = 64/3, which is not a whole number.

Therefore, 68 is not a term in the A.P.

(ii) Given, A.P. 3, 8, 13,…

Here, a = 3 and d = a₂ – a₁ = 8 – 3 = 5

We know that, the n^th term a_n = a + (n – 1)d

Required to check the n^th term a_n = 302

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n – 5 = 302

5n – 2 = 302

5n = 304

⇒ n = 304/5, which is not a whole number.

Therefore, 302 is not a term in the A.P.

(iii) Given, A.P. 11, 8, 5, 2, …

Here, a = 11 and d = a₂ – a₁ = 8 – 11 = -3

We know that, the n^th term a_n = a + (n – 1)d

Required to check the n^th term a_n = -150

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 – 3n + 3 = -150

3n = 150 + 14

3n = 164

⇒ n = 164/3, which is not a whole number.

Therefore, -150 is not a term in the A.P.

4. How many terms are there in the A.P.?

(i) 7, 10, 13, ….., 43

(ii) -1, -5/6, -2/3, -1/2, … , 10/3

(iii) 7, 13, 19, …, 205

(iv) 18, 15½, 13, …., -47

Solution:

(i) Given, A.P. 7, 10, 13, ….., 43

Here, a = 7 and d = a₂ – a₁ = 10 – 7 = 3

We know that, the n^th term a_n = a + (n – 1)d

And, given the n^th term a_n = 43

a + (n – 1)d = 43

7 + (n – 1)(3) = 43

7 + 3n – 3 = 43

3n = 43 – 4

3n = 39

⇒ n = 13

Therefore, there are 13 terms in the given A.P.

(ii) Given, A.P. -1, -5/6, -2/3, -1/2, … , 10/3

Here, a = -1 and d = a₂ – a₁ = -5/6 – (-1) = 1/6

We know that, the n^th term a_n = a + (n – 1)d

And, given the n^th term a_n = 10/3

a + (n – 1)d = 10/3

-1 + (n – 1)(1/6) = 10/3

-1 + n/6 – 1/6 = 10/3

n/6 = 10/3 + 1 + 1/6

n/6 = (20 + 6 + 1)/6

n = (20 + 6 + 1)

⇒ n = 27

Therefore, there are 27 terms in the given A.P.

(iii) Given, A.P. 7, 13, 19, …, 205

Here, a = 7 and d = a₂ – a₁ = 13 – 7 = 6

We know that, the n^th term a_n = a + (n – 1)d

And, given the n^th term a_n = 205

a + (n – 1)d = 205

7 + (n – 1)(6) = 205

7 + 6n – 6 = 205

6n = 205 – 1

n = 204/6

⇒ n = 34

Therefore, there are 34 terms in the given A.P.

(iv) Given, A.P. 18, 15½, 13, …., -47

Here, a = 7 and d = a₂ – a₁ = 15½ – 18 = 5/2

We know that, the n^th term a_n = a + (n – 1)d

And, given the n^th term a_n = -47

a + (n – 1)d = 43

18 + (n – 1)(-5/2) = -47

18 – 5n/2 + 5/2 = -47

36 – 5n + 5 = -94

5n = 94 + 36 + 5

5n = 135

⇒ n = 27

Therefore, there are 27 terms in the given A.P.

5. The first term of an A.P. is 5, the common difference is 3, and the last term is 80; find the number of terms.

Solution:

Given,

a = 5 and d = 3

We know that, the n^th term a_n = a + (n – 1)d

So, for the given A.P. a_n = 5 + (n – 1)3 = 3n + 2

Also given, last term = 80

⇒ 3n + 2 = 80

3n = 78

n = 78/3 = 26

Therefore, there are 26 terms in the A.P.

6. The 6^th and 17^th terms of an A.P. are 19 and 41, respectively. Find the 40^th term.

Solution:

Given,

a₆ = 19 and a₁₇ = 41

We know that, the n^th term a_n = a + (n – 1)d

So,

a₆ = a + (6-1)d

⇒ a + 5d = 19 …… (i)

Similarity,

a₁₇ = a + (17 – 1)d

⇒ a + 16d = 41 …… (ii)

Solving (i) and (ii),

(ii) – (i) ⇒

a + 16d – (a + 5d) = 41 – 19

11d = 22

⇒ d = 2

Using d in (i), we get

a + 5(2) = 19

a = 19 – 10 = 9

Now, the 40^th term is given by a₄₀ = 9 + (40 – 1)2 = 9 + 78 = 87

Therefore, the 40^th term is 87.

7. If the 9^th term of an A.P. is zero, prove its 29^th term is double the 19^th term.

Solution:

Given,

a₉ = 0

We know that, the n^th term a_n = a + (n – 1)d

So, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ……(i)

Now,

The 29^th term is given by a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

And, a₂₉ = (a + 8d) + 20d [using (i)]

⇒ a₂₉ = 20d ….. (ii)

Similarly, the 19^th term is given by a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

And, a₁₉ = (a + 8d) + 10d [using (i)]

⇒ a₁₉ = 10d …..(iii)

On comparing (ii) and (iii), it’s clearly seen that

a₂₉ = 2(a₁₉)

Therefore, the 29^th term is double the 19^th term.

8. If 10 times the 10^th term of an A.P. is equal to 15 times the 15^th term, show that the 25^th term of the A.P. is zero.

Solution:

Given,

10 times the 10^th term of an A.P. is equal to 15 times the 15^th term.

We know that, the n^th term a_n = a + (n – 1)d

⇒ 10(a₁₀) = 15(a₁₅)

10(a + (10 – 1)d) = 15(a + (15 – 1)d)

10(a + 9d) = 15(a + 14d)

10a + 90d = 15a + 210d

5a + 120d = 0

5(a + 24d) = 0

a + 24d = 0

a + (25 – 1)d = 0

⇒ a₂₅ = 0

Therefore, the 25^th term of the A.P. is zero.

9. The 10^th and 18^th terms of an A.P. are 41 and 73, respectively. Find the 26^th term.

Solution:

Given,

A₁₀ = 41 and a₁₈ = 73

We know that, the n^th term a_n = a + (n – 1)d

So,

a₁₀ = a + (10 – 1)d

⇒ a + 9d = 41 …… (i)

Similarity,

a₁₈ = a + (18 – 1)d

⇒ a + 17d = 73 …… (ii)

Solving (i) and (ii),

(ii) – (i) ⇒

a + 17d – (a + 9d) = 73 – 41

8d = 32

⇒ d = 4

Using d in (i), we get

a + 9(4) = 41

a = 41 – 36 = 5

Now, the 26^th term is given by a₂₆ = 5 + (26 – 1)4 = 5 + 100 = 105

Therefore, the 26^th term is 105.

10. In a certain A.P., the 24^th term is twice the 10^th term. Prove that the 72^nd term is twice the 34^th term.

Solution:

Given,

The 24^th term is twice the 10^th term.

We know that, the n^th term a_n = a + (n – 1)d

⇒ a₂₄ = 2(a₁₀)

a + (24 – 1)d = 2(a + (10 – 1)d)

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d …. (1)

Now, the 72^nd term can be expressed as

a₇₂ = a + (72 – 1)d

= a + 71d

= a + 5d + 66d

= a + a + 66d [using (1)]

= 2(a + 33d)

= 2(a + (34 – 1)d)

= 2(a₃₄)

⇒ a₇₂ = 2(a₃₄)

Hence, the 72^nd term is twice the 34^th term of the given A.P.

11. The 26^th, 11^th, and the last term of an A.P. are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.

Solution:

Given,

a₂₆ = 0 , a₁₁ = 3 and a_n (last term) = -1/5 of an A.P.

We know that, the n^th term a_n = a + (n – 1)d

Then,

a₂₆ = a + (26 – 1)d

⇒ a + 25d = 0 …..(1)

And,

a₁₁ = a + (11 – 1)d

⇒ a + 10d = 3 …… (2)

Solving (1) and (2),

(1) – (2) ⇒

a + 25d – (a + 10d) = 0 – 3

15d = -3

⇒ d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term a_n = -1/5

⇒ 5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 – n + 1 = -1

n = 27

Therefore, the A.P. has 27 terms, and its common difference is -1/5.

12. If the n^th term of the A.P. 9, 7, 5, …. is the same as the n^th term of the A.P. 15, 12, 9, … find n.

Solution:

Given,

A.P₁ = 9, 7, 5, …. and A.P₂ = 15, 12, 9, …

And, we know that, the n^th term a_n = a + (n – 1)d

For A.P₁,

a = 9, d = Second term – first term = 9 – 7 = -2

And, its the n^th term a_n = 9 + (n – 1)(-2) = 9 – 2n + 2

a_n = 11 – 2n …..(i)

Similarly, for A.P₂

a = 15, d = Second term – first term = 12 – 15 = -3

And, its the n^th term a_n = 15 + (n – 1)(-3) = 15 – 3n + 3

a_n = 18 – 3n …..(ii)

According to the question, it’s given that

The n^th term of the A.P₁ = n^th term of the A.P₂

⇒ 11 – 2n = 18 – 3n

n = 7

Therefore, the 7^th term of both the A.P.s is equal.

13. Find the 12^th term from the end of the following arithmetic progressions.

(i) 3, 5, 7, 9, …. 201

(ii) 3,8,13, … ,253

(iii) 1, 4, 7, 10, … ,88

Solution:

In order the find the 12^th term for the end of an A.P. which has n terms, it’s done by simply finding the ((n -12) + 1)^th of the A.P.

And we know, the n^th term a_n = a + (n – 1)d

(i) Given A.P = 3, 5, 7, 9, …. 201

Here, a = 3 and d = (5 – 3) = 2

Now, find the number of terms when the last term is known, i.e., 201.

a_n = 3 + (n – 1)2 = 201

3 + 2n – 2 = 201

2n = 200

n = 100

Hence, the A.P. has 100 terms.

So, the 12^th term from the end is the same as (100 – 12 + 1)^th of the A.P, which is the 89^th term.

⇒ a₈₉ = 3 + (89 – 1)2

= 3 + 88(2)

= 3 + 176

= 179

Therefore, the 12^th term from the end of the A.P. is 179.

(ii) Given A.P = 3,8,13, … ,253

Here, a = 3 and d = (8 – 3) = 5

Now, find the number of terms when the last term is known, i.e., 253.

a_n = 3 + (n – 1)5 = 253

3 + 5n – 5 = 253

5n = 253 + 2 = 255

n = 255/5

n = 51

Hence, the A.P. has 51 terms.

So, the 12^th term from the end is the same as (51 – 12 + 1)^th of the A.P., which is the 40^th term.

⇒ a₄₀ = 3 + (40 – 1)5

= 3 + 39(5)

= 3 + 195

= 198

Therefore, the 12^th term from the end of the A.P. is 198.

(iii) Given A.P = 1, 4, 7, 10, … ,88

Here, a = 1 and d = (4 – 1) = 3

Now, find the number of terms when the last term is known, i.e., 88

a_n = 1 + (n – 1)3 = 88

1 + 3n – 3 = 88

3n = 90

n = 30

Hence, the A.P. has 30 terms.

So, the 12^th term from the end is the same as (30 – 12 + 1)^th of the A.P, which is the 19^th term.

⇒ a₈₉ = 1 + (19 – 1)3

= 1 + 18(3)

= 1 + 54

= 55

Therefore, the 12^th term from the end of the A.P. is 55.

14. The 4^th term of an A.P. is three times the first, and the 7^th term exceeds twice the third term by 1. Find the first term and the common difference.

Solution:

Let’s consider the first term and the common difference of the A.P. to be a and d, respectively.

Then, we know that a_n = a + (n – 1)d

Given conditions,

4^th term of an A.P. is three times the first

Expressing this by equation, we have,

⇒ a₄ = 3(a)

a + (4 – 1)d = 3a

3d = 2a ⇒ a = 3d/2…….(i)

And,

The 7^th term exceeds twice the third term by 1.

⇒ a₇ = 2(a₃) + 1

a + (7 – 1)d = 2(a + (3–1)d) + 1

a + 6d = 2a + 4d + 1

a – 2d +1 = 0 ….. (ii)

Using (i) in (ii), we have

3d/2 – 2d + 1 = 0

3d – 4d + 2 = 0

d = 2

So, putting d = 2 in (i), we get a

⇒ a = 3

Therefore, the first term is 3, and the common difference is 2.

15. Find the second term, and the n^th term of an A.P. whose 6^th term is 12 and the 8^th term is 22.

Solution:

Given, in an A.P

a₆ = 12 and a₈ = 22

We know that a_n = a + (n – 1)d

So,

a₆ = a + (6-1)d = a + 5d = 12 …. (i)

And,

a₈ = a + (8-1)d = a + 7d = 22 ……. (ii)

Solving (i) and (ii), we have

(ii) – (i) ⇒

a + 7d – (a + 5d) = 22 – 12

2d = 10

d = 5

Putting d in (i), we get,

a + 5(5) = 12

a = 12 – 25

a = -13

Thus, for the A.P, a = -13 and d = 5

So, the n^th term is given by a_n = a + (n-1)d

a_n = -13 + (n-1)5 = -13 + 5n – 5

⇒ a_n = 5n – 18

Hence, the second term is given by a₂ = 5(2) – 18 = 10 – 18 = -8

16. How many numbers of two digits are divisible by 3?

Solution:

The first 2-digit number divisible by 3 is 12. And the last 2-digit number divisible by 3 is 99.

So, this forms an A.P.

12, 15, 18, 21, …. , 99

Where, a = 12 and d = 3

Finding the number of terms in this A.P.

⇒ 99 = 12 + (n-1)3

99 = 12 + 3n – 3

90 = 3n

n = 90/3 = 30

Therefore, there are 30 two-digit numbers divisible by 3.

17. An A.P. consists of 60 terms. If the first and the last terms are 7 and 125, respectively. Find the 32^nd term.

Solution:

Given, an A.P of 60 terms.

And, a = 7 and a₆₀ = 125

We know that a_n = a + (n – 1)d

⇒ a₆₀ = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

So, the 32^nd term is given by

a₃₂ = 7 + (32 -1)2 = 7 + 62 = 69

⇒ a₃₂ = 69

18. The sum of the 4^th and 8^th terms of an A.P. is 24, and the sum of the 6^th and 10^th terms is 34. Find the first term and the common difference of the A.P.

Solution:

Given, in an A.P

The sum of the 4^th and 8^th terms of an A.P. is 24.

⇒ a₄ + a₈ = 24

And, we know that a_n = a + (n – 1)d

2a + 10d = 24

a + 5d = 12 …. (i)

Also given that,

The sum of the 6^th and 10^th terms is 34.

⇒ a₆ + a₁₀ = 34

2a + 14d = 34

a + 7d = 17 …… (ii)

Subtracting (i) form (ii), we have

a + 7d – (a + 5d) = 17 – 12

2d = 5

d = 5/2

Using d in (i), we get,

a + 5(5/2) = 12

a = 12 – 25/2

a = -1/2

Therefore, the first term is -1/2, and the common difference is 5/2.

19. The first term of an A.P. is 5, and its 100^th term is -292. Find the 50^th term of this A.P.

Solution:

Given, an A.P whose

a = 5 and a₁₀₀ = -292

We know that a_n = a + (n – 1)d

a100 = 5 + 99d = -292

99d = -297

d = -3

Hence, the 50^th term is

a₅₀ = a + 49d = 5 + 49(-3) = 5 – 147 = -142

20. Find a₃₀ – a₂₀ for the A.P.

(i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, ……

Solution:

We know that a_n = a + (n – 1)d

So, a₃₀ – a₂₀ = (a + 29d) – (a + 19) =10d

(i) Given A.P. -9, -14, -19, -24

Here, a = -9 and d = -14 – (-9) = = -14 + 9 = -5

So, a₃₀ – a₂₀ = 10d

= 10(-5)

= -50

(ii) Given A.P. a, a+d, a+2d, a+3d, ……

So, a₃₀ – a₂₀ = (a + 29d) – (a + 19d)

=10d

21. Write the expression a_n – a_k for the A.P. a, a+d, a+2d, …..

Hence, find the common difference of the A.P. for which

(i) 11^th term is 5 and 13^th term is 79.

(ii) a₁₀ – a₅ = 200

(iii) 20^th term is 10 more than the 18^th term.

Solution:

Given A.P. a, a+d, a+2d, …..

So, a_n = a + (n-1)d = a + nd –d

And, a_k = a + (k-1)d = a + kd – d

a_n – a_k = (a + nd – d) – (a + kd – d)

= (n – k)d

(i) Given 11^th term is 5 and 13^th term is 79.

Here n = 13 and k = 11,

a₁₃ – a₁₁ = (13 – 11)d = 2d

⇒ 79 – 5 = 2d

d = 74/2 = 37

(ii) Given, a₁₀ – a₅ = 200

⇒ (10 – 5)d = 200

5d = 200

d = 40

(iii) Given, the 20^th term is 10 more than the 18^th term.

⇒ a₂₀ – a₁₈ = 10

(20 – 18)d = 10

2d = 10

d = 5

22. Find n if the given value of x is the n^th term of the given A.P.

(i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, …; x = -151

(iii) 5½, 11, 16½, 22, ….; x = 550 (iv) 1, 21/11, 31/11, 41/11, …; x = 171/11

Solution:

(i) Given A.P. 25, 50, 75, 100, …… ,1000

Here, a = 25 d = 50 – 25 = 25

Last term (n^th term) = 1000

We know that a_n = a + (n – 1)d

⇒ 1000 = 25 + (n-1)25

1000 = 25 + 25n – 25

n = 1000/25

n = 40

(ii) Given A.P. -1, -3, -5, -7, …., -151

Here, a = -1 d = -3 – (-1) = -2

Last term (n^th term) = -151

We know that a_n = a + (n – 1)d

⇒ -151 = -1 + (n-1)(-2)

-151 = -1 – 2n + 2

n = 152/2

n = 76

(iii) Given A.P. 5½, 11, 16½, 22, … , 550

Here, a = 5½ d = 11 – (5½) = 5½ = 11/2

Last term (n^th term) = 550

We know that a_n = a + (n – 1)d

⇒ 550 = 5½ + (n-1)(11/2)

550 x 2 = 11+ 11n – 11

1100 = 11n

n = 100

(iv) Given A.P. 1, 21/11, 31/11, 41/11, 171/11

Here, a = 1 d = 21/11 – 1 = 10/11

Last term (n^th term) = 171/11

We know that a_n = a + (n – 1)d

⇒ 171/11 = 1 + (n-1)10/11

171 = 11 + 10n – 10

n = 170/10

n = 17

23. The eighth term of an A.P. is half of its second term, and the eleventh term exceeds one-third of its fourth term by 1. Find the 15^th term.

Solution:

Given, an A.P in which,

a₈ = 1/2(a₂)

a₁₁ = 1/3(a₄) + 1

We know that a_n = a + (n – 1)d

⇒ a₈ = 1/2(a₂)

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 …… (i)

And, a₁₁ = 1/3(a₄) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 …… (ii)

Solving (i) and (ii), by (ii) – 2x(i) ⇒

2a + 27d – 2(a + 13d) = 3 – 0

d = 3

Putting d in (i), we get,

a + 13(3) = 0

a = -39

Thus, the 15^th term a₁₅ = -39 + 14(3) = -39 + 42 = 3

24. Find the arithmetic progression whose third term is 16, and the seventh term exceeds its fifth term by 12.

Solution:

Given, in an A.P

a₃ = 16 and a₇ = a₅ + 12

We know that a_n = a + (n – 1)d

⇒ a + 2d = 16…… (i)

And,

a + 6d = a + 4d + 12

2d = 12

⇒ d = 6

Using d in (i), we have

a + 2(6) = 16

a = 16 – 12 = 4

Hence, the A.P is 4, 10, 16, 22, …….

25. The 7^th term of an A.P. is 32, and its 13^th term is 62. Find the A.P.

Solution:

Given,

a₇ = 32 and a₁₃ = 62

From a_n – a_k = (a + nd – d) – (a + kd – d)

= (n – k)d

a₁₃ – a₇ = (13 – 7)d = 62 – 32 = 30

6d = 30

d = 5

Now,

a₇ = a + (7 – 1)5 = 32

a + 30 = 32

a = 2

Hence, the A.P is 2, 7, 12, 17, ……

26. Which term of the A.P. 3, 10, 17, …. will be 84 more than its 13^th term?

Solution:

Given, A.P. 3, 10, 17, ….

Here, a = 3 and d = 10 – 3 = 7

According to the question,

a_n = a₁₃ + 84

Using a_n = a + (n – 1)d,

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n – 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

n = 25

Therefore, it is the 25^th term, which is 84 more than its 13^th term.

27. Two arithmetic progressions have the same common difference. The difference between their 100^th terms is 100, what is the difference between their 1000^th terms?

Solution:

Let the two A.Ps be A.P₁ and A.P₂

For A.P_1, the first term = a and the common difference = d

And for A.P_2, the first term = b and the common difference = d

So, from the question we have

a₁₀₀ – b₁₀₀ = 100

(a + 99d) – (b + 99d) = 100

a – b = 100

Now, the difference between their 1000^th terms is,

(a + 999d) – (b + 999d) = a – b = 100

Therefore, the difference between their 1000^th terms is also 100.

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5 Page No: 9.30

1. Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P.
Solution:

Given,

(8x + 4), (6x – 2) and (2x + 7) are in A.P.

So, the common difference between the consecutive terms should be the same.

(6x – 2) – (8x + 4) = (2x + 7) – (6x – 2)

⇒ 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2

⇒ -2x – 6 = -4x + 9

⇒ -2x + 4x = 9 + 6

⇒ 2x = 15

Therefore, x = 15/2

2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Solution:

Given,

x + 1, 3x and 4x + 2 are in A.P.

So, the common difference between the consecutive terms should be the same.

3x – x – 1 = 4x + 2 – 3x

⇒ 2x – 1 = x + 2

⇒ 2x – x = 2 + 1

⇒ x = 3

Therefore, x = 3

3. Show that (a – b)², (a² + b²) and (a + b)² are in A.P.
Solution:

If (a – b)², (a² + b²) and (a + b)² have to be in A.P. then,

It should satisfy the condition,

2b = a + c [for a, b, c are in A.P]

Thus,

2 (a² + b²) = (a – b)² + (a + b)²

2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab

2 (a² + b²) = 2a² + 2b² = 2 (a² + b²)

LHS = RHS

Hence, proved.

4. The sum of three terms of an A.P. is 21, and the product of the first and the third terms exceeds the second term by 6, find three terms.
Solution:

Let’s consider the three terms of the A.P. to be a – d, a, a + d

So, the sum of three terms = 21

⇒ a – d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7

And, the product of the first and 3rd = 2nd term + 6

⇒ (a – d) (a + d) = a + 6

a² – d² = a + 6

⇒ (7 )² – d² = 7 + 6

⇒ 49 – d² = 13

⇒ d² = 49 – 13 = 36

⇒ d² = (6)²

⇒ d = 6

Hence, the terms are 7 – 6, 7, 7 + 6 ⇒ 1, 7, 13

5. Three numbers are in A.P. If the sum of these numbers is 27 and the product 648, find the numbers.
Solution:

Let the three numbers of the A.P. be a – d, a, a + d

From the question,

Sum of these numbers = 27

a – d + a + a + d = 27

⇒ 3a = 27

a = 27/3 = 9

Now, the product of these numbers = 648

(a – d)(a)(a + d) = 648

a(a² – d²) = 648

a² – 648/a = d²

9² – (648/9) = d²

9³ – 648 = 9d²

729 – 648 = 9d²

81 = 9d²

d² = 9

d = 3 or -3

Hence, the terms are 9-3, 9 and 9+3 ⇒ 6, 9, 12 or 12, 9, 6 (for d = -3)

6. Find the four numbers in A.P. whose sum is 50 and in which the greatest number is 4 times the least.
Solution:

Let’s consider the four terms of the A.P. to be (a – 3d), (a – d), (a + d) and (a + 3d).

From the question,

Sum of these terms = 50

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50

⇒ a – 3d + a – d + a + d + a – 3d= 50

⇒ 4a = 50

⇒ a = 50/4 = 25/2

And, also given that the greatest number = 4 x the least number

⇒ a + 3d = 4 (a – 3d)

⇒ a + 3d = 4a – 12d

⇒ 4a – a = 3d + 12d

⇒3a = 15d

⇒a = 5d

Using the value of a in the above equation, we have

⇒25/2 = 5d

⇒ d = 5/2

So, the terms will be

(a – 3d) = (25/2 – 3(5/2)), (a – d) = (25/2 – 5/2), (25/2 + 5/2) and (25/2 + 3(5/2)).

⇒ 5, 10, 15, 20

RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.6 Page No: 9.50

1. Find the sum of the following arithmetic progressions.

(i) 50, 46, 42, … to 10 terms

(ii) 1, 3, 5, 7, … to 12 terms

(iii) 3, 9/2, 6, 15/2, … to 25 terms

(iv) 41, 36, 31, … to 12 terms

(v) a + b, a – b, a – 3b, … to 22 terms

(vi) (x – y)², (x² + y²), (x + y)², to 22 tams

(viii) – 26, – 24, – 22, …. to 36 terms

Solution:

In an A.P., if the first term = a, common difference = d, and if there are n terms.

Then, the sum of n terms is given by:

(i) Given A.P.is 50, 46, 42 to 10 term.

First term (a) = 50

Common difference (d) = 46 – 50 = – 4

n^th term (n) = 10

= 5{100 – 9.4}

= 5{100 – 36}

= 5 × 64

∴ S₁₀ = 320

(ii) Given A.P is 1, 3, 5, 7, …..to 12 terms.

First term (a) = 1

Common difference (d) = 3 – 1 = 2

n^th term (n) = 12

= 6 × {2 + 22} = 6.24

∴ S₁₂ = 144

(iii) Given A.P. is 3, 9/2, 6, 15/2, … to 25 terms

First term (a) = 3

Common difference (d) = 9/2 – 3 = 3/2

Sum of n terms S_n, given n = 25

(iv) Given expression is 41, 36, 31, ….. to 12 terms.

First term (a) = 41

Common difference (d) = 36 – 41 = -5

Sum of n terms S_n, given n = 12

(v) a + b, a – b, a – 3b, ….. to 22 terms

First term (a) = a + b

Common difference (d) = a – b – a – b = -2b

Sum of n terms S_n = n/2{2a(n – 1). d}

Here n = 22

S₂₂ = 22/2{2.(a + b) + (22 – 1). -2b}

= 11{2(a + b) – 22b)

= 11{2a – 20b}

= 22a – 440b

∴S₂₂ = 22a – 440b

(vi) (x – y)²,(x² + y²), (x + y)²,… to n terms

First term (a) = (x – y)²

Common difference (d) = x² + y² – (x – y)²

= x² + y² – (x² + y² – 2xy)

= x² + y² – x² + y² + 2xy

= 2xy

Sum of n^th terms S_n = n/2{2a + (n – 1). d}

= n/2{2(x – y)² + (n – 1). 2xy}

= n{(x – y)² + (n – 1)xy}

∴ S_n = n{(x — y)² + (n — 1). xy)

(viii) Given expression -26, – 24. -22, to 36 terms

First term (a) = -26

Common difference (d) = -24 – (-26)

= -24 + 26 = 2

Sum of n terms, S_n = n/2{2a + (n – 1)d) for n = 36

S_n = 36/2{2(-26) + (36 – 1)2}

= 18[-52 + 70]

= 18×18

= 324

∴ S_n = 324

2. Find the sum to n terms of the A.P. 5, 2, –1, – 4, –7, …

Solution:

Given AP is 5, 2, -1, -4, -7, …..

Here, a = 5, d = 2 – 5 = -3

We know that,

S_n = n/2{2a + (n – 1)d}

= n/2{2.5 + (n – 1) – 3}

= n/2{10 – 3(n – 1)}

= n/2{13 – 3n)

∴ S_n = n/2(13 – 3n)

3. Find the sum of n terms of an A.P. whose terms are given by a_n = 5 – 6n.

Solution:

Given the nth term of the A.P as a_n = 5 – 6n

Put n = 1, and we get

a₁ = 5 – 6.1 = -1

So, the first term (a) = -1

Last term (a_n) = 5 – 6n = 1

Then, S_n = n/2(-1 + 5 – 6n)

= n/2(4 – 6n) = n(2 – 3n)

4. Find the sum of the last ten terms of the A.P.: 8, 10, 12, 14, .. , 126

Solution:

Given A.P. 8, 10, 12, 14, .. , 126

Here, a = 8 , d = 10 – 8 = 2

We know that, a_n = a + (n – 1)d

So, to find the number of terms,

126 = 8 + (n – 1)2

126 = 8 + 2n – 2

2n = 120

n = 60

Next, let’s find the 51^st term.

a₅₁ = 8 + 50(2) = 108

So, the sum of the last ten terms is the sum of a₅₁ + a₅₂ + a₅₃ + ……. + a₆₀

Here, n = 10, a = 108 and l = 126

S = 10/2 [108 + 126]

= 5(234)

= 1170

Hence, the sum of the last ten terms of the A.P. is 1170.

5. Find the sum of the first 15 terms of each of the following sequences having the n^th term as:

(i) a_n = 3 + 4n 

(ii) b_n = 5 + 2n 

(iii) x_n = 6 – n 

(iv) y_n = 9 – 5n

Solution:

(i) Given an A.P. whose n^th term is given by a_n = 3 + 4n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 3 + 4(1) = 3 + 4 = 7

For the last term (l), here, n = 15

a₁₅ = 3 + 4(15) = 63

So, S_n = 15(7 + 63)/2

= 15 x 35

= 525

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 525

(ii) Given an A.P. whose n^th term is given by b_n = 5 + 2n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given b_n, we get

a = 5 + 2(1) = 5 + 2 = 7

For the last term (l), here, n = 15

a₁₅ = 5 + 2(15) = 35

So, S_n = 15(7 + 35)/2

= 15 x 21

= 315

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 315

(iii) Given an A.P. whose n^th term is given by x_n = 6 – n

To find the sum of the n terms of the given A.P., using the formula

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given x_n, we get

a = 6 – 1 = 5

For the last term (l), here n = 15

a₁₅ = 6 – 15 = -9

So, S_n = 15(5 – 9)/2

= 15 x (-2)

= -30

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = -30

(iv) Given an A.P. whose n^th term is given by y_n = 9 – 5n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given y_n, we get

a = 9 – 5(1) = 9 – 5 = 4

For the last term (l), here n = 15

a₁₅ = 9 – 5(15) = -66

So, S_n = 15(4 – 66)/2

= 15 x (-31)

= -465

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = -465

6. Find the sum of the first 20 terms of the sequence whose n^th term is a_n = An + B.

Solution:

Given an A.P. whose nth term is given by, a_n = An + B

We need to find the sum of the first 20 terms.

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term,

Putting n = 1 in the given a_n, we get

a = A(1) + B = A + B

For the last term (l), here n = 20

A₂₀ = A(20) + B = 20A + B

S₂₀ = 20/2((A + B) + 20A + B)

= 10[21A + 2B]

= 210A + 20B

Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B

7. Find the sum of the first 25 terms of an A.P. whose n^th term is given by a_n = 2 – 3n.

Solution:

Given an A.P. whose n^th term is given by a_n = 2 – 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 2 – 3(1) = -1

For the last term (l), here n = 25

a₂₅ = 2 – 3(25) = -73

So, S_n = 25(-1 – 73)/2

= 25 x (-37)

= -925

Therefore, the sum of the 25 terms of the given A.P. is S₂₅ = -925

8. Find the sum of the first 25 terms of an A.P. whose n^th term is given by a_n = 7 – 3n. 

Solution:

Given an A.P. whose n^th term is given by a_n = 7 – 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 7 – 3(1) = 7 – 3 = 4

For the last term (l), here n = 25

a₁₅ = 7 – 3(25) = -68

So, S_n = 25(4 – 68)/2

= 25 x (-32)

= -800

Therefore, the sum of the 15 terms of the given A.P. is S₂₅ = -800

9. If the sum of a certain number of terms starting from the first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.

Solution:

Given the sum of the certain number of terms of an A.P. = 116

We know that, S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms So for the given A.P.(25, 22, 19,…)

Here, we have the first term (a) = 25

The sum of n terms S_n = 116

Common difference of the A.P. (d) = a₂ – a₁ = 22 – 25 = -3

Now, substituting values in S_n

⟹ 116 = n/2[2(25) + (n − 1)(−3)]

⟹ (n/2)[50 + (−3n + 3)]  = 116

⟹ (n/2)[53 − 3n] = 116

⟹ 53n – 3n² = 116 x 2

Thus, we get the following quadratic equation,

3n² – 53n + 232 = 0

By factorisation method of solving, we have

⟹ 3n² – 24n – 29n + 232 = 0

⟹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

⟹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

⟹ n = 29/3

Also, n – 8 = 0

⟹ n = 8

Since n cannot be a fraction, the number of terms is taken as 8.

So, the term is:

a₈ = a₁ + 7d = 25 + 7(-3) = 25 – 21 = 4

Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4.

10. (i) How many terms of the sequence 18, 16, 14….  should be taken so that their sum is zero. 

(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2, respectively, and the sum of the terms is 40? 

(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636? 

(iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? 

(v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? 

Solution:

(i) Given AP. is 18, 16, 14, …

We know that,

S_n = n/2[2a + (n − 1)d]

Here,

The first term (a) = 18

The sum of n terms (S_n) = 0 (given)

Common difference of the A.P.

(d) = a₂  – a₁ = 16 – 18 = – 2

So, on substituting the values in S_n

⟹ 0 = n/2[2(18) + (n − 1)(−2)]

⟹ 0 = n/2[36 + (−2n + 2)]

⟹ 0 = n/2[38 − 2n] Further, n/2

⟹ n = 0 Or, 38 – 2n = 0

⟹ 2n = 38

⟹ n = 19

The number of terms cannot be zero. Hence, the number of terms (n) should be 19.

(ii) Given, the first term (a) = -14, Filth term (a₅) = 2, Sum of terms (S_n) = 40 of the A.P.

If the common difference is taken as d.

Then, a₅ = a + 4d

⟹ 2 = -14 + 4d

⟹ 2 + 14 = 4d

⟹ 4d = 16

⟹ d = 4

Next, we know that S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Now, on substituting the values in S_n

⟹ 40 = n/2[2(−14) + (n − 1)(4)]

⟹ 40 = n/2[−28 + (4n − 4)]

⟹ 40 = n/2[−32 + 4n]

⟹ 40(2) = – 32n + 4n²

So, we get the following quadratic equation,

4n² – 32n – 80 = 0

⟹ n² – 8n – 20 = 0

On solving by factorisation method, we get

n² – 10n + 2n – 20 = 0

⟹ n(n – 10) + 2( n – 10 ) = 0

⟹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

⟹ n = -2

Or, n – 10 = 0

⟹ n = 10

Since the number of terms cannot be negative.

Therefore, the number of terms (n) is 10.

(iii) Given AP is 9, 17, 25,…

We know that,

S_n = n/2[2a + (n − 1)d]

Here we have,

The first term (a) = 9 and the sum of n terms (S_n) = 636

Common difference of the A.P. (d) = a₂  – a₁ = 17 – 9 = 8

Substituting the values in S_n, we get

⟹ 636 = n/2[2(9) + (n − 1)(8)]

⟹ 636 = n/2[18 + (8n − 8)]

⟹ 636(2) = (n)[10 + 8n]

⟹ 1271 = 10n + 8n²

Now, we get the following quadratic equation,

⟹ 8n² + 10n – 1272 = 0

⟹ 4n²+ 5n – 636 = 0

On solving by factorisation method, we have

⟹ 4n² – 48n + 53n – 636 = 0

⟹ 4n(n – 12) + 53(n – 12) = 0

⟹ (4n + 53)(n – 12) = 0

Either 4n + 53 = 0 ⟹ n = -53/4

Or, n – 12 = 0 ⟹ n = 12

The number of terms cannot be a fraction.

Therefore, the number of terms (n) is 12.

(iv) Given A.P. is 63, 60, 57,…

We know that,

S_n = n/2[2a + (n − 1)d]

Here we have,

the first term (a) = 63

The sum of n terms (S_n) = 693

Common difference of the A.P. (d) = a₂ – a₁ = 60 – 63 = –3

On substituting the values in S_n, we get

⟹ 693 = n/2[2(63) + (n − 1)(−3)]

⟹ 693 = n/2[126+(−3n + 3)]

⟹ 693 = n/2[129 − 3n]

⟹ 693(2) = 129n – 3n²

Now, we get the following quadratic equation.

⟹ 3n² – 129n + 1386 = 0

⟹ n² – 43n + 462

Solving by factorisation method, we have

⟹ n² – 22n – 21n + 462 = 0

⟹ n(n – 22) -21(n – 22) = 0

⟹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 ⟹ n = 22

Or,  n – 21 = 0 ⟹ n = 21

Now, the 22^nd term will be a₂₂ = a₁ + 21d = 63 + 21( -3 ) = 63 – 63 = 0

So, the sum of 22, as well as 21 terms, is 693.

Therefore, the number of terms (n) is 21 or 22.

(v) Given A.P. is 27, 24, 21. . .

We know that,

S_n = n/2[2a + (n − 1)d]

Here, we have the first term (a) = 27

The sum of n terms (S_n) = 0

Common difference of the A.P. (d) = a₂ – a₁ = 24 – 27 = -3

On substituting the values in S_n, we get

⟹ 0 = n/2[2(27) + (n − 1)( − 3)]

⟹ 0 = (n)[54 + (n – 1)(-3)]

⟹ 0 = (n)[54 – 3n + 3]

⟹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0

⟹ 3n = 57

⟹ n = 19

The number of terms cannot be zero,

Hence, the number of terms (n) is 19.

11. Find the sum of the first

(i) 11 terms of the A.P. : 2, 6, 10, 14,  . . . 

(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . . 

(iii) 51 terms of the A.P. : whose second term is 2 and the fourth term is 8.

Solution:

We know that the sum of terms for different arithmetic progressions is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given A.P 2, 6, 10, 14,… to 11 terms.

Common difference (d) = a₂ – a₁ = 10 – 6 = 4

Number of terms (n) = 11

The first term for the given A.P. (a) = 2

So,

S₁₁  = 11/2[2(2) + (11 − 1)4]

= 11/2[2(2) + (10)4]

= 11/2[4 + 40]

= 11 × 22

= 242

Hence, the sum of the first 11 terms for the given A.P. is 242.

(ii) Given A.P. – 6, 0, 6, 12, … to 13 terms.

Common difference (d) = a₂ – a₁ = 6 – 0 = 6

Number of terms (n) = 13

First term (a) = -6

So,

S₁₃  = 13/2[2(− 6) + (13 –1)6]

= 13/2[(−12) + (12)6]

= 13/2[60] = 390

Hence, the sum of the first 13 terms for the given A.P. is 390.

(iii) 51 terms of an AP whose a₂ = 2 and a₄ = 8

We know that, a₂ = a + d

2 = a + d  …(2)

Also, a₄ = a + 3d

8 = a + 3d  … (2)

Subtracting (1) from (2), we have

2d = 6

d = 3

Substituting d = 3 in (1), we get

2 = a + 3

⟹ a = -1

Given that the number of terms (n) = 51

First term (a) = -1

So,

S_n  = 51/2[2(−1) + (51 − 1)(3)]

= 51/2[−2 + 150]

= 51/2[148]

= 3774

Hence, the sum of the first 51 terms for the A.P. is 3774.

12. Find the sum of

(i) the first 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 – digit natural numbers which are divisible by 13.

(iv) all 3 – digit natural numbers which are multiples of 11.

Solution:

We know that the sum of terms for an A.P is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given, the first 15 multiples of 8.

These multiples form an A.P: 8, 16, 24, …… , 120

Here, a = 8 , d = 61 – 8 = 8 and the number of terms(n) = 15

Now, finding the sum of 15 terms, we have

Hence, the sum of the first 15 multiples of 8 is 960

(ii)(a) First 40 positive integers divisible by 3.

Hence, the first multiple is 3, and the 40^th multiple is 120.

And, these terms will form an A.P. with a common difference of 3.

Here, the first term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

So, the sum of 40 terms

S₄₀  = 40/2[2(3) + (40 − 1)3]

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123) = 2460

Thus, the sum of the first 40 multiples of 3 is 2460.

(b) First 40 positive integers divisible by 5

Hence, the first multiple is 5, and the 40^th multiple is 200.

And these terms will form an A.P. with a common difference of 5.

Here, the first term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

So, the sum of 40 terms

S₄₀  = 40/2[2(5) + (40 − 1)5]

= 20[10 + (39)5]

= 20 (10 + 195)

= 20 (205) = 4100

Hence, the sum of the first 40 multiples of 5 is 4100.

(c) First 40 positive integers divisible by 6

Hence, the first multiple is 6, and the 40^th multiple is 240.

And these terms will form an A.P. with a common difference of 6.

Here, the first term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

So, the sum of 40 terms

S₄₀  = 40/2[2(6) + (40 − 1)6]

= 20[12 + (39)6]

=20(12 + 234)

= 20(246) = 4920

Hence, the sum of the first 40 multiples of 6 is 4920.

(iii) All 3-digit natural numbers which are divisible by 13.

So, we know that the first 3-digit multiple of 13 is 104, and the last 3-digit multiple of 13 is 988.

And these terms form an A.P. with a common difference of 13.

Here, first term (a) = 104 and the last term (l) = 988

Common difference (d) = 13

Finding the number of terms in the A.P. by a_n = a + (n − 1)d

We have,

988 = 104 + (n – 1)13

⟹ 988 = 104 + 13n -13

⟹ 988 = 91 + 13n

⟹ 13n = 897

⟹ n = 69

Now, using the formula for the sum of n terms, we get

S₆₉  = 69/2[2(104) + (69 − 1)13]

= 69/2[208 + 884]

= 69/2[1092]

= 69(546)

= 37674

Hence, the sum of all 3-digit multiples of 13 is 37674.

(iv) All 3-digit natural numbers which are multiples of 11.

So, we know that the first 3-digit multiple of 11 is 110, and the last 3-digit multiple of 13 is 990.

And, these terms form an A.P. with a common difference of 11.

Here, first term (a) = 110 and the last term (l) = 990

Common difference (d) = 11

Finding the number of terms in the A.P. by a_n = a + (n − 1)d

We get,

990 = 110 + (n – 1)11

⟹ 990 = 110 + 11n -11

⟹ 990 = 99 + 11n

⟹ 11n = 891

⟹ n = 81

Now, using the formula for the sum of n terms, we get

S₈₁  = 81/2[2(110) + (81 − 1)11]

= 81/2[220 + 880]

= 81/2[1100]

= 81(550)

= 44550

Hence, the sum of all 3-digit multiples of 11 is 44550.

13. Find the sum:

(i) 2 + 4 + 6 + . . . + 200 

(ii) 3 + 11 + 19 + . . . + 803 

(iii) (-5) + (-8) + (-11) + . . . + (- 230) 

(iv) 1 + 3 + 5 + 7 + . . . + 199 

(vi) 34 + 32 + 30 + . . . + 10 

(vii) 25 + 28 + 31 + . . . + 100 

Solution:

We know that the sum of terms for an A.P is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = n/2[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

(i) Given series. 2 + 4 + 6 + . . . + 200 which is an A.P

Where, a = 2 ,d = 4 – 2 = 2 and last term (a_n = l) = 200

We know that, a_n = a + (n – 1)d

So,

200 = 2 + (n – 1)2

200 = 2 + 2n – 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S₁₀₀ = 100/2 [2 + 200]

= 50(202)

= 10100

Hence, the sum of the terms of the given series is 10100.

(ii) Given series. 3 + 11 + 19 + . . . + 803 which is an A.P

Where, a = 3 ,d = 11 – 3 = 8 and last term (a_n = l) = 803

We know that, a_n = a + (n – 1)d

So,

803 = 3 + (n – 1)8

803 = 3 + 8n – 8

n = 808/8 = 101

Now, for the sum of these 101 terms

S₁₀₁ = 101/2 [3 + 803]

= 101(806)/2

= 101 x 403

= 40703

Hence, the sum of the terms of the given series is 40703.

(iii) Given series (-5) + (-8) + (-11) + . . . + (- 230) which is an A.P

Where, a = -5 ,d = -8 – (-5) = -3 and last term (a_n = l) = -230

We know that, a_n = a + (n – 1)d

So,

-230 = -5 + (n – 1)(-3)

-230 = -5 – 3n + 3

3n = -2 + 230

n = 228/3 = 76

Now, for the sum of these 76 terms

S₇₆ = 76/2 [-5 + (-230)]

= 38 x (-235)

= -8930

Hence, the sum of the terms of the given series is -8930.

(iv) Given series. 1 + 3 + 5 + 7 + . . . + 199 which is an A.P

Where, a = 1 ,d = 3 – 1 = 2 and last term (a_n = l) = 199

We know that, a_n = a + (n – 1)d

So,

199 = 1 + (n – 1)2

199 = 1 + 2n – 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S₁₀₀ = 100/2 [1 + 199]

= 50(200)

= 10000

Hence, the sum of the terms of the given series is 10000.

(v) Given series, which is an A.P.

Where, a = 7, d = 10 ½ – 7 = (21 – 14)/2 = 7/2 and last term (a_n = l) = 84

We know that, a_n = a + (n – 1)d

So,

84 = 7 + (n – 1)(7/2)

168 = 14 + 7n – 7

n = (168 – 7)/7 = 161/7 = 23

Now, for the sum of these 23 terms

S₂₃ = 23/2 [7 + 84]

= 23(91)/2

= 2093/2

Hence, the sum of the terms of the given series is 2093/2.

(vi) Given series, 34 + 32 + 30 + . . . + 10 which is an A.P

Where, a = 34 ,d = 32 – 34 = -2 and last term (a_n = l) = 10

We know that, a_n = a + (n – 1)d

So,

10 = 34 + (n – 1)(-2)

10 = 34 – 2n + 2

n = (36 – 10)/2 = 13

Now, for the sum of these 13 terms,

S₁₃ = 13/2 [34 + 10]

= 13(44)/2

= 13 x 22

= 286

Hence, the sum of the terms of the given series is 286.

(vii) Given series, 25 + 28 + 31 + . . . + 100 which is an A.P

Where, a = 25 ,d = 28 – 25 = 3 and last term (a_n = l) = 100

We know that, a_n = a + (n – 1)d

So,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

n = (100 – 22)/3 = 26

Now, for the sum of these 26 terms,

S₁₀₀ = 26/2 [25 + 100]

= 13(125)

= 1625

Hence, the sum of the terms of the given series is 1625.

14. The first and the last terms of an A.P. are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?

Solution:

Given, the first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference (d) of the A.P. = 9

Let the number of terms be n. And, we know that l = a + (n – 1)d

So, 350 = 17 + (n- 1) 9

⟹ 350 = 17 + 9n – 9

⟹ 350 = 8 + 9n

⟹ 350 – 8 = 9n

Thus, we get, n = 38

Now, finding the sum of terms,

S_n  = n/2[a + l]

= 38/2(17 + 350)

= 19 × 367

= 6973

Hence, the number of terms of the A.P. is 38, and their sum is 6973.

15. The third term of an A.P. is 7, and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of the first 20 terms.

Solution:

Let’s consider the first term as a and the common difference as d.

Given,

a₃ = 7 …. (1) and,

a₇ = 3a₃ + 2   …. (2)

So, using (1) in (2), we get,

a₇ = 3(7) + 2 = 21 + 2 = 23  …. (3)

Also, we know that

a_n = a +(n – 1)d

So, the 3th term (for n = 3),

a₃ = a + (3 – 1)d

⟹ 7 = a + 2d   (Using 1)

⟹ a = 7 – 2d     …. (4)

Similarly, for the 7th term (n = 7),

a₇ = a + (7 – 1) d 24 = a + 6d = 23  (Using 3)

a = 23 – 6d  …. (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

⟹ 0 = 23 – 6d – 7 + 2d

⟹ 0 = 16 – 4d

⟹ 4d = 16

⟹ d = 4

Now, to find a, we substitute the value of d in  (4), a =7 – 2(4)

⟹ a = 7 – 8

a = -1

Hence, for the A.P. a = -1 and d = 4

For finding the sum, we know that

S_n = n/2[2a + (n − 1)d] and n = 20 (given)

S₂₀  = 20/2[2(−1) + (20 − 1)(4)]

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

= (10)[74]

= 740

Hence, the sum of the first 20 terms for the given A.P. is 740

16. The first term of an A.P. is 2, and the last term is 50. The sum of all these terms is 442. Find the common difference.

Solution:

Given,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 50

Sum of all the terms S_n = 442

So, let the common difference of the A.P. be taken as d.

The sum of all the terms is given as,

442 = (n/2)(2 + 50)

⟹ 442 = (n/2)(52)

⟹ 26n = 442

⟹ n = 17

Now, the last term is expressed as

50 = 2 + (17 – 1)d

⟹ 50 = 2 + 16d

⟹ 16d = 48

⟹ d = 3

Thus, the common difference of the A.P. is d = 3.

17. If the 12^th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?

Solution:

Let us take the first term as a and the common difference as d.

Given,

a₁₂ = -13 S₄ = 24

Also, we know that a_n = a + (n – 1)d

So, for the 12th term

a₁₂ = a + (12 – 1)d = -13

⟹ a + 11d = -13

a = -13 – 11d  …. (1)

And, we know that for the sum of terms is

S_n = n/2[2a + (n − 1)d]

Here, n = 4

S₄  = 4/2[2(a) + (4 − 1)d]

⟹ 24 = (2)[2a + (3)(d)]

⟹ 24 = 4a + 6d

⟹ 4a = 24 – 6d

Subtracting (1) from (2), we have

Further simplifying for d, we get,

⟹ -19 × 2 = 19d

⟹ d = – 2

On substituting the value of d in (1), we find a

a = -13 – 11(-2)

a = -13 + 22

a = 9

Next, the sum of 10 terms is given by

S₁₀  = 10/2[2(9) + (10 − 1)(−2)]

= (5)[19 + (9)(-2)]

= (5)(18 – 18) = 0

Thus, the sum of the first 10 terms for the given A.P. is S₁₀ = 0.

18.

19. In an A.P., if the first term is 22, the common difference is – 4, and the sum to n terms is 64, find n.

Solution:

Given that,

a = 22, d = – 4 and S_n = 64

Let us consider the number of terms as n.

For the sum of terms in an A.P, we know that

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

So,

⟹ S_n = n/2[2(22) + (n − 1)(−4)]

⟹ 64 = n/2[2(22) + (n − 1)(−4)]

⟹ 64(2) = n(48 – 4n)

⟹ 128 = 48n – 4n²

After rearranging the terms, we have a quadratic equation

4n² – 48n + 128 = 0,

n² – 12n + 32 = 0 [dividing by 4 on both sides]

n² – 12n + 32 = 0

Solving by factorisation method,

n² – 8n – 4n + 32 = 0

n ( n – 8 ) – 4 ( n – 8 ) = 0

(n – 8) (n – 4) = 0

So, we get n – 8 = 0 ⟹ n = 8

Or, n – 4 = 0 ⟹ n = 4

Hence, the number of terms can be either n = 4 or 8.

20. In an A.P., if the 5^th and 12^th terms are 30 and 65, respectively, what is the sum of the first 20 terms? 

Solution:

Let’s take the first term as a and the common difference to be d.

Given that,

a₅ = 30  and a₁₂ = 65

And, we know that a_n = a + (n – 1)d

So,

a₅ = a + (5 – 1)d

30 = a + 4d

a = 30 – 4d   …. (i)

Similarly, a₁₂ = a + (12 – 1) d

65 = a + 11d

a = 65 – 11d …. (ii)

Subtracting (i) from (ii), we have

a – a = (65 – 11d) – (30 – 4d)

0 = 65 – 11d – 30 + 4d

0 = 35 – 7d

7d = 35

d = 5

Putting d in (i), we get

a = 30 – 4(5)

a = 30 – 20

a = 10

Thus for the A.P; d = 5 and a = 10

Next, to find the sum of the first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.

S_n = n/2[2a + (n − 1)d]

Where;

a = First term of the given A.P.

d = Common difference of the given A.P.

n = Number of terms

Here, n = 20, so we have

S₂₀ = 20/2[2(10) + (20 − 1)(5)]

= (10)[20 + (19)(5)]

= (10)[20 + 95]

= (10)[115]

= 1150

Hence, the sum of the first 20 terms for the given A.P. is 1150

21. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.

Solution:

Let’s take the first term as a and the common difference as d.

Given that,

a₂ = 14 and a₃ = 18

And, we know that a_n = a + (n – 1)d

So,

a₂ = a + (2 – 1)d

⟹ 14 = a + d

⟹ a = 14 – d …. (i)

Similarly,

a₃ = a + (3 – 1)d

⟹ 18 = a + 2d

⟹ a = 18 – 2d …. (ii)

Subtracting (i) from (ii), we have

a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Putting d in (i) to find a,

a = 14 – 4

a = 10

Thus, for the A.P. d = 4 and a = 10

Now, to find the sum of terms,

S_n = n/2(2a + (n − 1)d)

Where,

a = The first term of the A.P.

d = Common difference of the A.P.

n = Number of terms So, using the formula for

n = 51,

⟹ S₅₁  = 51/2[2(10) + (51 – 1)(4)]

= 51/2[20 + (40)4]

= 51/2[220]

= 51(110)

= 5610

Hence, the sum of the first 51 terms of the given A.P. is 5610.

22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

Solution:

Given,

The sum of 7 terms of an A.P. is 49

⟹ S₇ = 49

And, the sum of 17 terms of an A.P. is 289

⟹ S₁₇ = 289

Let the first term of the A.P. be a and common difference as d.

And, we know that the sum of n terms of an A.P is

S_n = n/2[2a + (n − 1)d]

So,

S₇ = 49 = 7/2[2a + (7 – 1)d]

= 7/2 [2a + 6d]

= 7[a + 3d]

⟹ 7a + 21d = 49

a + 3d = 7 ….. (i)

Similarly,

S₁₇ = 17/2[2a + (17 – 1)d]

= 17/2 [2a + 16d]

= 17[a + 8d]

⟹ 17[a + 8d] = 289

a + 8d = 17 ….. (ii)

Now, subtracting (i) from (ii), we have

a + 8d – (a + 3d) = 17 – 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 – 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_n  = n/2[2(1) + (n − 1)(2)]

= n/2[2 + 2n – 2]

= n/2[2n]

= n²

Therefore, the sum of n terms of the A.P. is given by n².

23. The first term of an A.P. is 5, the last term is 45, and the sum is 400. Find the number of terms and the common difference.

Solution:

The sum of first n terms of an A.P is given by S_n = n/2(2a + (n − 1)d)

Given,

First term (a) = 5, last term (a_n) = 45 and sum of n terms (S_n) = 400

Now, we know that

a_n = a + (n – 1)d

⟹ 45 = 5 + (n – 1)d

⟹ 40 = nd – d

⟹ nd – d = 40 …. (1)

Also,

S_n = n/2(2(a) + (n − 1)d)

400 = n/2(2(5) + (n − 1)d)

800 = n (10 + nd – d)

800 = n (10 + 40) [using (1)]

⟹ n = 16

Putting n in (1), we find d

nd – d = 40

16d – d = 40

15d = 40

d = 8/3

Therefore, the common difference of the given A.P. is 8/3.

24. In an A.P., the first term is 8, the n^th term is 33, and the sum of the first n term is 123. Find n and d, the common difference. 

Solution:

Given,

The first term of the A.P (a) = 8

The nth term of the A.P (l) = 33

And the sum of all the terms S_n = 123

Let the common difference of the A.P. be d.

So, find the number of terms by

123 = (n/2)(8 + 33)

123 = (n/2)(41)

n = (123 x 2)/ 41

n = 246/41

n = 6

Next, to find the common difference between the A.P., we know that

l = a + (n – 1)d

33 = 8 + (6 – 1)d

33 = 8 + 5d

5d = 25

d = 5

Thus, the number of terms is n = 6, and the common difference of the A.P. is d = 5.

25. In an A.P., the first term is 22, the n^th term is -11, and the sum of the first n term is 66. Find n and d, the common difference. 

Solution:

Given,

The first term of the A.P (a) = 22

The nth term of the A.P (l) = -11

And the sum of all the terms S_n = 66

Let the common difference of the A.P. be d.

So, finding the number of terms by

66 = (n/2)[22 + (−11)]

66 = (n/2)[22 − 11]

(66)(2) = n(11)

6 × 2 = n

n = 12

Now, to find d,

We know that, l = a + (n – 1)d

– 11 = 22 + (12 – 1)d

-11 = 22 + 11d

11d = – 33

d = – 3

Hence, the number of terms is n = 12 and the common difference d = -3

26. The first and the last terms of an A.P. are 7 and 49, respectively. If the sum of all its terms is 420, find the common difference.

Solution:

Given,

First term (a) = 7, last term (a_n) = 49 and sum of n terms (S_n) = 420

Now, we know that

a_n = a + (n – 1)d

⟹ 49 = 7 + (n – 1)d

⟹ 43 = nd – d

⟹ nd – d = 42  ….. (1)

Next,

S_n = n/2(2(7) + (n − 1)d)

⟹ 840 = n[14 + nd – d]

⟹ 840 = n[14 + 42] [using (1)]

⟹ 840 = 54n

⟹ n = 15 …. (2)

So, by substituting (2) in (1), we have

nd – d = 42

⟹ 15d – d = 42

⟹ 14d = 42

⟹ d = 3

Therefore, the common difference of the given A.P. is 3.

27. The first and the last terms of an A.P. are 5 and 45, respectively. If the sum of all its terms is 400, find its common difference.

Solution:

Given,

First term (a) = 5 and the last term (l) = 45

Also, S_n = 400

We know that,

a_n = a + (n – 1)d

⟹ 45 = 5 + (n – 1)d

⟹ 40 = nd – d

⟹ nd – d = 40  ….. (1)

Next,

S_n = n/2(2(5) + (n − 1)d)

⟹ 400 = n[10 + nd – d]

⟹ 800 = n[10 + 40] [using (1)]

⟹ 800 = 50n

⟹ n = 16 …. (2)

So, by substituting (2) in (1), we have

nd – d = 40

⟹ 16d – d = 40

⟹ 15d = 40

⟹ d = 8/3

Therefore, the common difference of the given A.P. is 8/3.

28. The sum of the first q terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1: 2. Find the first and 15^th terms of the A.P.

Solution:

Let a be the first term and d be the common difference.

And we know that the sum of the first n terms is

S_n = n/2(2a + (n − 1)d)

Also, nth term is given by a_n = a + (n – 1)d

From the question, we have

S_q = 162 and a₆ : a₁₃ = 1 : 2

So,

2a₆ = a₁₃

⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d

⟹ 2a + 10d = a + 12d

⟹ a = 2d  …. (1)

And, S₉ = 162

⟹ S₉ = 9/2(2a + (9 − 1)d)

⟹ 162 = 9/2(2a + 8d)

⟹ 162 × 2 = 9[4d + 8d]  [from (1)]

⟹ 324 = 9 × 12d

⟹ d = 3

⟹ a = 2(3) [from (1)]

⟹ a = 6

Hence, the first term of the A.P. is 6.

For the 15^th term, a₁₅ = a + 14d = 6 + 14 × 3 = 6 + 42

Therefore, a₁₅ = 48

29. If the 10^th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^th term.

Solution:

Let’s consider a to be the first term and d be the common difference.

And we know that the sum of the first n terms is

S_n = n/2(2a + (n − 1)d) and n^th term is given by: a_n = a + (n – 1)d

Now, from the question we have

S₁₀ = 120

⟹ 120 = 10/2(2a + (10 − 1)d)

⟹ 120 = 5(2a + 9d)

⟹ 24 = 2a + 9d …. (1)

Also, given that, a₁₀ = 21

⟹ 21 = a + (10 – 1)d

⟹ 21 = a + 9d …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

⟹a = 3

Now, by putting a = 3 in equation (2), we get

3 + 9d = 21

9d = 18

d = 2

Thus, we have the first term(a) = 3 and the common difference(d) = 2

Therefore, the n^th term is given by

a_n  = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

= 2n + 1

Hence, the n^th term of the A.P is (a_n) = 2n + 1.

30. The sum of the first 7 terms of an A.P. is 63, and the sum of its next 7 terms is 161. Find the 28^th term of this A.P.

Solution:

Let’s take a to be the first term and d to be the common difference.

And we know that sum of the first n terms

S_n = n/2(2a + (n − 1)d)

Given that sum of the first 7 terms of an A.P. is 63.

S₇ = 63

And the sum of the next 7 terms is 161.

So, the sum of the first 14 terms = Sum of the first 7 terms + sum of the next 7 terms

S₁₄ = 63 + 161 = 224

Now, having

S₇ = 7/2(2a + (7 − 1)d)

⟹ 63(2) = 7(2a + 6d)

⟹ 9 × 2 = 2a + 6d

⟹ 2a + 6d = 18 . . . . (1)

And,

S₁₄ = 14/2(2a + (14 − 1)d)

⟹ 224 = 7(2a + 13d)

⟹ 32 = 2a + 13d …. (2)

Now, subtracting (1) from (2), we get

⟹ 13d – 6d = 32 – 18

⟹ 7d = 14

⟹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from the n^th term

⟹ a₂₈  = a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

Therefore, the 28^th term is 57.

31. The sum of the first seven terms of an A.P. is 182. If its 4^th and 17^th terms are in a ratio 1: 5, find the A.P.

Solution:

Given that,

S₁₇ = 182

And, we know that the sum of the first n term is

S_n = n/2(2a + (n − 1)d)

So,

S₇ = 7/2(2a + (7 − 1)d)

182 × 2 = 7(2a + 6d)

364 = 14a + 42d

26 = a + 3d

a = 26 – 3d … (1)

Also, it’s given that the 4^th term and 17^th term are in a ratio of 1: 5. So, we have

⟹ 5(a₄) = 1(a₁₇)

⟹ 5 (a + 3d) = 1 (a + 16d)

⟹ 5a + 15d = a + 16d

⟹ 4a = d …. (2)

Now, substituting (2) in (1), we get

⟹ 4 ( 26 – 3d ) = d

⟹ 104 – 12d = d

⟹ 104 = 13d

⟹ d = 8

Putting d in (2), we get

⟹ 4a = d

⟹ 4a = 8

⟹ a = 2

Therefore, the first term is 2, and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . ..

32. The n^th term of an A.P. is given by (-4n + 15). Find the sum of the first 20 terms of this A.P.

Solution:

Given,

The n^th term of the A.P = (-4n + 15)

So, by putting n = 1 and n = 20, we can find the first ans 20^th term of the A.P

a = (-4(1) + 15) = 11

And,

a₂₀ = (-4(20) + 15) = -65

Now, to find the sum of 20 terms of this A.P. we have the first and last term.

So, using the formula

S_n = n/2(a + l)

S₂₀  = 20/2(11 + (-65))

= 10(-54)

= -540

Therefore, the sum of the first 20 terms of this A.P. is -540.

33. In an A.P., the sum of the first ten terms is -150, and the sum of its next 10 terms is -550. Find the A.P.

Solution:

Let’s take a to be the first term and d to be the common difference.

And we know that the sum of the first n terms,

S_n = n/2(2a + (n − 1)d)

Given that sum of the first 10 terms of an A.P. is -150.

S₁₀ = -150

And the sum of the next 10 terms is -550.

So, the sum of the first 20 terms = Sum of the first 10 terms + sum of the next 10 terms

S₂₀ = -150 + -550 = -700

Now, having

S₁₀ = 10/2(2a + (10 − 1)d)

⟹ -150 = 5(2a + 9d)

⟹ -30 = 2a + 9d

⟹ 2a + 9d = -30 . . . . (1)

And,

S₂₀ = 20/2(2a + (20 − 1)d)

⟹ -700 = 10(2a + 19d)

⟹ -70 = 2a + 19d …. (2)

Now, subtracting (1) from (2), we get

⟹ 19d – 9d = -70 – (-30)

⟹ 10d = -40

⟹ d = -4

Using d in (1), we have

2a + 9(-4) = -30

2a = -30 + 36

a = 6/2 = 3

Hence, we have a = 3 and d = -4

So, the A.P is 3, -1, -5, -9, -13,…..

34. Sum of the first 14 terms of an A.P. is 1505, and its first term is 10. Find its 25^th term.

Solution

Given,

The first term of the A.P is 1505 and

S₁₄ = 1505

We know that the sum of the first n terms is

S_n = n/2(2a + (n − 1)d)

So,

S₁₄ = 14/2(2(10) + (14 − 1)d) = 1505

7(20 + 13d) = 1505

20 + 13d = 215

13d = 215 – 20

d = 195/13

d =15

Thus, the 25^th term is given by

a₂₅ = 10 + (25 -1)15

= 10 + (24)15

= 10 + 360

= 370

Therefore, the 25^th term of the A.P is 370

35. In an A.P., the first term is 2, the last term is 29, and the sum of the terms is 155. Find the common difference between the A.P.

Solution:

Given,

The first term of the A.P. (a) = 2

The last term of the A.P. (l) = 29

And the sum of all the terms (S_n) = 155

Let the common difference of the A.P. be d.

So, find the number of terms by the sum of terms formula.

S_n = n/2 (a + l)

155 = n/2(2 + 29)

155(2) = n(31)

31n = 310

n = 10

Using n for the last term, we have

l = a + (n – 1)d

29 = 2 + (10 – 1)d

29 = 2 + (9)d

29 – 2 = 9d

9d = 27

d = 3

Hence, the common difference of the A.P. is d = 3

36. The first and the last term of an A.P. are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?

Solution:

Given,

In an A.P first term (a) = 17 and the last term (l) = 350

And the common difference (d) = 9

We know that,

a_n = a + (n – 1)d

so,

a_n = l = 17 + (n – 1)9 = 350

17 + 9n – 9 = 350

9n = 350 – 8

n = 342/9

n = 38

So, the sum of all the terms of the A.P is given by

S_n = n/2 (a + l)

= 38/2(17 + 350)

= 19(367)

= 6973

Therefore, the sum of the terms of the A.P. is 6973.

37. Find the number of terms of the A.P. –12, –9, –6, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

Solution:

Given,

The first term, a = -12

Common difference, d = a₂ – a₁ = – 9 – (- 12)

d = – 9 + 12 = 3

And, we know that n^th term = a_n = a + (n – 1)d

⟹ 21 = -12 + (n – 1)3

⟹ 21 = -12 + 3n – 3

⟹ 21 = 3n – 15

⟹ 36 = 3n

⟹  n = 12

Thus, the number of terms is 12.

Now, if 1 is added to each of the 12 terms, the sum will increase by 12.

Hence, the sum of all the terms of the A.P. so obtained is

⟹ S₁₂ + 12  = 12/2[a + l] + 12

= 6[-12 + 21] + 12

= 6 × 9 + 12

= 66

Therefore, the sum after adding 1 to each of the terms in the A.P. is 66.