Arithmetic progressions is a high weightage chapter in Class 10 Maths. This exercise deals with problems related to the sum of terms of an A.P. Numerous real-time problems are discussed under this exercise and the RD Sharma Solutions Class 10 is the best tool for students to refer and prepare well for their exams. This exercise solutions of RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.6 PDF is given below.

## RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.6 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.6

**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}, (x^{2}Â + y^{2}), (x + y)^{2}, to 22 tams**

**(viii) â€“ 26, â€“ 24, â€“ 22, …. to 36 terms**

**Solution:**

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

Then, 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 = â€“ 4

n^{thÂ }term (n) = 10

= 5{100 – 9.4}

= 5{100 – 36}

= 5 Ã— 64

âˆ´ S_{10}Â = 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}Â term (n) = 12

= 6 Ã— {2 + 22} = 6.24

âˆ´ S_{12}Â = 144

(iii) Given A.P. isÂ 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, â€¦.. to 12 terms.

First term (a) = 41

Common difference (d) = 36 – 41 = -5

Sum of nÂ terms S_{n}, given n = 12

(v) a + b, a â€“ b, a – 3b, â€¦.. to 22 terms

First term (a) = a + b

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

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

Here n = 22

S_{22}Â = 22/2{2.(a + b) + (22 – 1). -2b}

= 11{2(a + b) – 22b)

= 11{2a – 20b}

= 22a – 440b

âˆ´S_{22}Â = 22a – 440b

(vi) (x – y)^{2},(x^{2}Â + y^{2}), (x + y)^{2},… to n terms

First term (a) = (x – y)^{2}

Common difference (d) = x^{2}Â + y^{2}Â – (x – y)^{2}

= x^{2}Â + y^{2}Â – (x^{2}Â + y^{2}Â – 2xy)

= x^{2}Â + y^{2}Â – x^{2}Â + y^{2}Â + 2xy

= 2xy

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

= n/2{2(x – y)^{2}Â + (n – 1). 2xy}

= n{(x – y)^{2}Â + (n – 1)xy}

âˆ´ S_{n}Â = n{(x â€” y)^{2}Â + (n â€” 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}Â = n/2{2a + (n – 1)d) for n = 36

S_{n}Â = 36/2{2(-26) + (36 – 1)2}

= 18[-52 + 70]

= 18×18

= 324

âˆ´ S_{n}Â = 324

**2. Find the sum to n terms of the A.P. 5, 2, â€“1, â€“ 4, â€“7, …**

**Solution:**

Given AP is 5, 2, -1, -4, -7, …..

Here, a = 5, d = 2 – 5 = -3

We know that,

S_{n}Â = n/2{2a + (n – 1)d}

= n/2{2.5 + (n – 1) – 3}

= n/2{10 – 3(n – 1)}

= n/2{13 – 3n)

âˆ´ S_{n}Â = n/2(13 – 3n)

**3. Find the sum of n terms of an A.P. whose the terms is given by a _{n} = 5 – 6n.**

**Solution:**

Given nth term of the A.P as a_{n}Â = 5 – 6n

Put n = 1, we get

a_{1}Â = 5 – 6.1 = -1

So, first term (a) = -1

Last term (a_{n}) = 5 – 6n = 1

Then, S_{n}Â = n/2(-1 + 5 – 6n)

= n/2(4 – 6n) = n(2 – 3n)

**4. Find the sum of 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 â€“ 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, letâ€™s find the 51^{st} term

a_{51} = 8 + 50(2) = 108

So, the sum of last ten terms is the sum of a_{51} + a_{52} + a_{53} + â€¦â€¦. + a_{60}

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

S = 10/2 [108 + 126]

= 5(234)

= 1170

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

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

**(i) a _{n}Â = 3 + 4nÂ **

**(ii) b _{n}Â = 5 + 2nÂ **

**(iii) x _{n}Â = 6 – nÂ **

**(iv) y _{n}Â = 9 â€“ 5n**

**Solution:**

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

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

S_{n}Â = 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

a_{15} = 3 + 4(15) = 63

So,Â S_{n}Â = 15(7 + 63)/2

= 15 x 35

= 525

Therefore, the sum of the 15 terms of the given A.P. is S_{15}Â = 525

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

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

S_{n}Â = 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

a_{15} = 5 + 2(15) = 35

So,Â S_{n}Â = 15(7 + 35)/2

= 15 x 21

= 315

Therefore, the sum of the 15 terms of the given A.P. is S_{15}Â = 315

(iii) Given an A.P. whose n^{th} term is given by x_{n}Â = 6 – n

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

S_{n}Â = 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 â€“ 1 = 5

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

a_{15} = 6 â€“ 15 = -9

So,Â S_{n}Â = 15(5 â€“ 9)/2

= 15 x (-2)

= -30

Therefore, the sum of the 15 terms of the given A.P. is S_{15}Â = -30

(iv) Given an A.P. whose n^{th} term is given by y_{n}Â = 9 â€“ 5n

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

S_{n}Â = 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 â€“ 5 = 4

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

a_{15} = 9 – 5(15) = -66

So,Â S_{n}Â = 15(4 – 66)/2

= 15 x (-31)

= -465

Therefore, the sum of the 15 terms of the given A.P. is S_{15}Â = -465

**6. Find the sum of first 20 terms the sequence whose n ^{th}Â term is a_{n}Â = An + B.**

**Solution:
**

Given an A.P. whose nth term is given by, a_{n}Â = An + B

We need to find the sum of first 20 terms.

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

S_{n}Â = 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

A_{20} = A(20) + B = 20A + B

S_{20}Â = 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 first 25 terms of an A.P whose n ^{th}Â term is given by a_{n}Â = 2 – 3n.**

**Solution:**

Given an A.P. whose n^{th} term is given by a_{n}Â = 2 â€“ 3n

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

S_{n}Â = 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

a_{25} = 2 – 3(25) = -73

So,Â S_{n}Â = 25(-1 – 73)/2

= 25 x (-37)

= -925

Therefore, the sum of the 25 terms of the given A.P. is S_{25}Â = -925

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

**Solution:**

Given an A.P. whose n^{th} term is given by a_{n}Â = 7 â€“ 3n

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

S_{n}Â = 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 â€“ 3 = 4

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

a_{15} = 7 – 3(25) = -68

So,Â S_{n}Â = 25(4 – 68)/2

= 25 x (-32)

= -800

Therefore, the sum of the 15 terms of the given A.P. is S_{25}Â = -800

**9. If the sum of a certain number of terms starting from 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}Â = n/2[2a + (n âˆ’ 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}Â = 116

Common difference of the A.P. (d) = a_{2}Â – a_{1}Â = 22 â€“ 25 = -3

Now, substituting values in S_{n}

âŸ¹Â 116 = n/2[2(25) + (n âˆ’ 1)(âˆ’3)]

âŸ¹Â (n/2)[50 + (âˆ’3n + 3)]Â = 116

âŸ¹Â (n/2)[53 âˆ’ 3n]Â = 116

âŸ¹ 53n – 3n^{2} = 116 x 2

Thus, we get the following quadratic equation,

3n^{2}Â – 53n + 232 = 0

By factorization method of solving, we have

âŸ¹ 3n^{2}Â – 24n – 29n + 232 = 0

âŸ¹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

âŸ¹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

âŸ¹ n =Â 29/3

Also, n – 8 = 0

âŸ¹ n = 8

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

So, the term is:

a_{8}Â = a_{1}Â + 7d = 25 + 7(-3) = 25 – 21 =Â 4

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

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

**(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?Â **

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

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

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

**Solution: **

(i) Given AP. is 18, 16, 14, …

We know that,

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Here,

The first term (a) = 18

The sum of n terms (S_{n}) = 0 (given)

Common difference of the A.P.

(d) = a_{2}Â Â – a_{1}Â = 16 – 18 = â€“ 2

So, on substituting the values in S_{n}

âŸ¹Â 0 = n/2[2(18) + (n âˆ’ 1)(âˆ’2)]

âŸ¹Â 0 = n/2[36 + (âˆ’2n + 2)]

âŸ¹Â 0 = n/2[38 âˆ’ 2n]Â Further,Â n/2

âŸ¹ n = 0 Or, 38 – 2n = 0

âŸ¹ 2n = 38

âŸ¹ n = 19

Since, the number of terms cannot be zer0, hence the number of terms (n) should be 19.

(ii) Given, the first term (a) = -14, Filth term (a_{5}) = 2, Sum of terms (S_{n}) = 40 of the A.P.

If the common difference is taken as d.

Then, a_{5}Â = a_{Â }+ 4d

âŸ¹ 2 = -14 + 4d

âŸ¹ 2 + 14 = 4d

âŸ¹ 4d = 16

âŸ¹ d = 4

Next, we know that S_{nÂ }= n/2[2a + (n âˆ’ 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}

âŸ¹Â 40 = n/2[2(âˆ’14) + (n âˆ’ 1)(4)]

âŸ¹Â 40 = n/2[âˆ’28 + (4n âˆ’ 4)]

âŸ¹Â 40 = n/2[âˆ’32 + 4n]

âŸ¹ 40(2) = – 32n + 4n^{2}

So, we get the following quadratic equation,

4n^{2}Â – 32n – 80 = 0

âŸ¹ n^{2}Â – 8n – 20 = 0

On solving by factorization method, we get

n^{2}Â – 10n + 2n – 20 = 0

âŸ¹ n(n – 10) + 2(n – 10 ) = 0

âŸ¹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

âŸ¹ n = -2

Or, n – 10 = 0

âŸ¹ 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}Â = n/2[2a + (n âˆ’ 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) = a_{2}Â Â – a_{1}Â = 17 – 9 = 8

Substituting the values in S_{n}, we get

âŸ¹Â 636 = n/2[2(9) + (n âˆ’ 1)(8)]

âŸ¹Â 636 = n/2[18 + (8n âˆ’ 8)]

âŸ¹ 636(2) = (n)[10 + 8n]

âŸ¹ 1271 = 10n + 8n^{2}

Now, we get the following quadratic equation,

âŸ¹ 8n^{2}Â + 10n – 1272 = 0

âŸ¹ 4n^{2}+ 5n – 636 = 0

On solving by factorisation method, we have

âŸ¹ 4n^{2}Â – 48n + 53n – 636 = 0

âŸ¹ 4n(n – 12) + 53(n – 12) = 0

âŸ¹ (4n + 53)(n – 12) = 0

Either 4n + 53 = 0 âŸ¹Â n = -53/4

Or, n – 12 = 0 âŸ¹ n = 12

Since, 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}Â = n/2[2a + (n âˆ’ 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) = a_{2}Â – a_{1}Â = 60 – 63 = â€“3

On substituting the values in S_{n }we get

âŸ¹Â 693 = n/2[2(63) + (n âˆ’ 1)(âˆ’3)]

âŸ¹Â 693 = n/2[126+(âˆ’3n + 3)]

âŸ¹Â 693 = n/2[129 âˆ’ 3n]

âŸ¹ 693(2) = 129n – 3n^{2}

Now, we get the following quadratic equation.

âŸ¹ 3n^{2}Â – 129n + 1386 = 0

âŸ¹ n^{2}Â – 43n + 462

Solving by factorisation method, we have

âŸ¹ n^{2}Â – 22n – 21n + 462 = 0

âŸ¹ n(n – 22) -21(n – 22) = 0

âŸ¹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 âŸ¹ n = 22

Or,Â n – 21 = 0 âŸ¹ n = 21

Now, the 22^{nd}Â term will be a_{22}Â = a_{1}Â + 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}Â = n/2[2a + (n âˆ’ 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) = a_{2}Â – a_{1}Â = 24 – 27 = -3

On substituting the values in S_{n}, we get

âŸ¹Â 0 = n/2[2(27) + (n âˆ’ 1)( âˆ’ 3)]

âŸ¹ 0 = (n)[54 + (n – 1)(-3)]

âŸ¹ 0 = (n)[54 – 3n + 3]

âŸ¹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0

âŸ¹ 3n = 57

âŸ¹ n = 19

The number of terms cannot be zero,

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

**11. Find the sum of the first**

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

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

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

**Solution:**

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

S_{n}Â = n/2[2a + (n âˆ’ 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) = a_{2}Â – a_{1}Â = 10 – 6 = 4

Number of terms (n) = 11

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

So,

S_{11}Â = 11/2[2(2) + (11 âˆ’ 1)4]

=Â 11/2[2(2) + (10)4]

=Â 11/2[4 + 40]

= 11 Ã— 22

= 242

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

(ii) Given A.P. â€“ 6, 0, 6, 12, … to 13 terms.

Common difference (d) = a_{2}Â – a_{1}Â = 6 – 0 = 6

Number of terms (n) = 13

First term (a) = -6

So,

S_{13}Â = 13/2[2(âˆ’ 6) + (13 â€“1)6]

=Â 13/2[(âˆ’12) + (12)6]

=Â 13/2[60]Â = 390

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

(iii) 51 terms of an AP whose a_{2}Â = 2 and a_{4}Â = 8

We know that, a_{2}Â = a + d

2 = a + dÂ …(2)

Also, a_{4}Â = a + 3d

8 = a + 3dÂ Â … (2)

Subtracting (1) from (2), we have

2d = 6

d = 3

Substituting d = 3 in (1), we get

2 = a + 3

âŸ¹ a = -1

Given that the number of terms (n) = 51

First term (a) = -1

So,

S_{n}Â = 51/2[2(âˆ’1) + (51 âˆ’ 1)(3)]

=Â 51/2[âˆ’2 + 150]

=Â 51/2[148]

= 3774

Hence, the sum of 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}Â = n/2[2a + (n âˆ’ 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, first 15 multiples of 8.

These multiples form an A.P: 8, 16, 24, â€¦â€¦ , 120

Here, a = 8 , d = 61 â€“ 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 the common difference of 3.

Here, First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

So, the sum of 40 terms

S_{40}Â = 40/2[2(3) + (40 âˆ’ 1)3]

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123) = 2460

Thus, the sum of 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 the common difference of 5.

Here, First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

So, the sum of 40 terms

S_{40 Â }= 40/2[2(5) + (40 âˆ’ 1)5]

= 20[10 + (39)5]

= 20 (10 + 195)

= 20 (205) = 4100

Hence, the sum of 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 the common difference of 6.

Here, First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

So, the sum of 40 terms

S_{40}Â = 40/2[2(6) + (40 âˆ’ 1)6]

= 20[12 + (39)6]

=20(12 + 234)

= 20(246) = 4920

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

(iii) All 3 digit natural number 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 the 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}Â = a + (n âˆ’ 1)d

We have,

988 = 104 + (n – 1)13

âŸ¹ 988 = 104 + 13n -13

âŸ¹ 988 = 91 + 13n

âŸ¹ 13n = 897

âŸ¹ n = 69

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

S_{69}Â = 69/2[2(104) + (69 âˆ’ 1)13]

=Â 69/2[208 + 884]

=Â 69/2[1092]

= 69(546)

= 37674

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

(iv) All 3 digit natural number 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 the 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}Â = a + (n âˆ’ 1)d

We get,

990 = 110 + (n – 1)11

âŸ¹ 990 = 110 + 11n -11

âŸ¹ 990 = 99 + 11n

âŸ¹ 11n = 891

âŸ¹ n = 81

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

S_{81}Â = 81/2[2(110) + (81 âˆ’ 1)11]

=Â 81/2[220 + 880]

=Â 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 + . . . + 200Â **

**(ii) 3 + 11 + 19 + . . . + 803Â **

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

**(iv) 1 + 3 + 5 + 7 + . . . + 199Â **

**(vi) 34 + 32 + 30 + . . . + 10Â **

**(vii) 25 + 28 + 31 + . . . + 100Â **

**Solution:**

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

S_{n}Â = n/2[2a + (n âˆ’ 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}Â = 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 + . . . + 200Â which is an A.P

Where, a = 2 ,d = 4 â€“ 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 â€“ 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S_{100 }= 100/2 [2 + 200]

= 50(202)

= 10100

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

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

Where, a = 3 ,d = 11 â€“ 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 â€“ 8

n = 808/8 = 101

Now, for the sum of these 101 terms

S_{101 }= 101/2 [3 + 803]

= 101(806)/2

= 101 x 403

= 40703

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

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

Where, a = -5 ,d = -8 â€“ (-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

S_{76 }= 76/2 [-5 + (-230)]

= 38 x (-235)

= -8930

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

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

Where, a = 1 ,d = 3 â€“ 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 â€“ 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S_{100 }= 100/2 [1 + 199]

= 50(200)

= 10000

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

(v) Given series which is an A.P

Where, a = 7, d = 10 Â½ – 7 = (21 â€“ 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 â€“ 7

n = (168 â€“ 7)/7 = 161/7 = 23

Now, for the sum of these 23 terms

S_{23 }= 23/2 [7 + 84]

= 23(91)/2

= 2093/2

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

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

Where, a = 34 ,d = 32 â€“ 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 â€“ 10)/2 = 13

Now, for the sum of these 13 terms

S_{13 }= 13/2 [34 + 10]

= 13(44)/2

= 13 x 22

= 286

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

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

Where, a = 25 ,d = 28 â€“ 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 â€“ 22)/3 = 26

Now, for the sum of these 26 terms

S_{100 }= 26/2 [25 + 100]

= 13(125)

= 1625

Hence, the sum of 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

âŸ¹ 350 = 17 + 9n – 9

âŸ¹ 350 = 8 + 9n

âŸ¹ 350 – 8 = 9n

Thus we get, n = 38

Now, finding the sum of terms

S_{n}Â = n/2[a + l]

= 38/2(17 + 350)

= 19 Ã— 367

= 6973

Hence, the number of terms is 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 first 20 terms.**

**Solution:**

Letâ€™s consider the first term as a and the common difference as d.

Given,

a_{3}Â = 7Â …. (1) and,

a_{7}Â = 3a_{3}Â + 2Â Â Â …. (2)

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

a_{7Â }= 3(7) + 2 = 21 + 2 = 23 Â …. (3)

Also, we know that

a_{n}Â = a +(n – 1)d

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

a_{3}Â = a + (3 – 1)d

âŸ¹ 7 = a + 2dÂ Â (Using 1)

âŸ¹ a = 7 – 2d Â Â Â Â …. (4)

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

a_{7}Â = a + (7 – 1) d 24 = a + 6dÂ = 23Â (Using 3)

a = 23 – 6d Â …. (5)

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

a – a = (23 – 6d) – (7 – 2d)

âŸ¹ 0 = 23 – 6d – 7 + 2d

âŸ¹ 0 = 16 – 4d

âŸ¹ 4d = 16

âŸ¹ d = 4

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

âŸ¹ a = 7 – 8

a = -1

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

For finding the sum, we know that

S_{n}Â = n/2[2a + (n âˆ’ 1)d]Â and n = 20 (given)

S_{20}Â = 20/2[2(âˆ’1) + (20 âˆ’ 1)(4)]

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

= (10)[-2 + 76]

= (10)[74]

= 740

Hence, the sum of 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)

âŸ¹Â 442 = (n/2)(52)

âŸ¹ 26n = 442

âŸ¹ n = 17

Now, the last term is expressed as

50 = 2 + (17 – 1)d

âŸ¹ 50 = 2 + 16d

âŸ¹ 16d = 48

âŸ¹ d = 3

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

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

**Solution:**

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

Given,

a_{12}Â = -13 S_{4}Â = 24

Also, we know that a_{nÂ }= a + (n – 1)d

So, for the 12th term

a_{12}Â = a + (12 – 1)d = -13

âŸ¹ a + 11d = -13

a = -13 – 11d Â …. (1)

And, we that for sum of terms

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Here, n = 4

S_{4}Â = 4/2[2(a) + (4 âˆ’ 1)d]

âŸ¹ 24 = (2)[2a + (3)(d)]

âŸ¹ 24 = 4a + 6d

âŸ¹ 4a = 24 – 6d

Subtracting (1) from (2), we have

Further simplifying for d, we get,

âŸ¹ -19 Ã— 2 = 19d

âŸ¹ d = â€“ 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 term is given by

S_{10}Â = 10/2[2(9) + (10 âˆ’ 1)(âˆ’2)]

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

= (5)(18 – 18) = 0

Thus, the sum of first 10 terms for the given A.P. is S_{10}Â = 0.

**18.**

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

**Solution:**

Given that,

a = 22, d = â€“ 4 and S_{n} = 64

Let us consider the number of terms as n.

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

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

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

So,

âŸ¹Â S_{n}Â = n/2[2(22) + (n âˆ’ 1)(âˆ’4)]

âŸ¹Â 64 = n/2[2(22) + (n âˆ’ 1)(âˆ’4)]

âŸ¹ 64(2) = n(48 – 4n)

âŸ¹ 128 = 48n – 4n^{2}

After rearranging the terms, we have a quadratic equation

4n^{2}Â – 48n + 128 = 0,

n^{2}Â – 12n + 32 = 0 [dividing by 4 on both sides]

n^{2}Â – 12n + 32 = 0

Solving by factorisation method,

n^{2}Â – 8n – 4n + 32 = 0

n ( n – 8 ) – 4 ( n – 8 ) = 0

(n – 8) (n – 4) = 0

So, we get n – 8 = 0 âŸ¹ n = 8

Or, n – 4 = 0 âŸ¹ n = 4

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

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

**Solution:**

Letâ€™s take the first term as a and the common difference to be d

Given that,

a_{5}Â = 30 Â and a_{12}Â = 65

And, we know that a_{n}Â = a + (n – 1)d

So,

a_{5}Â = a + (5 – 1)d

30 = a + 4d

a = 30 – 4dÂ Â …. (i)

Similarly, a_{12}Â = 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 first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

S_{n}Â = n/2[2a + (n âˆ’ 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

S_{20 }= 20/2[2(10) + (20 âˆ’ 1)(5)]

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

= (10)[20 + 95]

= (10)[115]

= 1150

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

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

**Solution:**

Letâ€™s take the first term as a and the common difference as d.

Given that,

a_{2}Â = 14 and a_{3}Â = 18

And, we know that a_{n}Â = a + (n – 1)d

So,

a_{2}Â = a + (2 – 1)d

âŸ¹ 14 = a + d

âŸ¹ a = 14 – d …. (i)

Similarly,

a_{3}Â = a + (3 – 1)d

âŸ¹ 18 = a + 2d

âŸ¹ 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 sum of terms

S_{n}Â = n/2(2a + (n âˆ’ 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,

âŸ¹ S_{51}Â = 51/2[2(10) + (51 – 1)(4)]

=Â 51/2[20 + (40)4]

=Â 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,

Sum of 7 terms of an A.P. is 49

âŸ¹ S_{7} = 49

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

âŸ¹ S_{17} = 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}Â = n/2[2a + (n âˆ’ 1)d]

So,

S_{7 }= 49 = 7/2[2a + (7 – 1)d]

= 7/2 [2a + 6d]

= 7[a + 3d]

âŸ¹ 7a + 21d = 49

a + 3d = 7 â€¦.. (i)

Similarly,

S_{17 }= 17/2[2a + (17 – 1)d]

= 17/2 [2a + 16d]

= 17[a + 8d]

âŸ¹ 17[a + 8d] = 289

a + 8d = 17 â€¦.. (ii)

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

a + 8d â€“ (a + 3d) = 17 â€“ 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 â€“ 6 = 1

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

For the sum of n terms is given by,

S_{n}Â = n/2[2(1) + (n âˆ’ 1)(2)]

= n/2[2 + 2n – 2]

= n/2[2n]

= n^{2}

Therefore, the sum of n terms of the A.P is given by n^{2}.

**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:**

Sum of first n terms of an A.P is given by S_{n}Â = n/2(2a + (n âˆ’ 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}Â = a + (n – 1)d

âŸ¹ 45 = 5 + (n – 1)d

âŸ¹ 40 = nd – d

âŸ¹ nd – d = 40 …. (1)

Also,

S_{nÂ }= n/2(2(a) + (n âˆ’ 1)d)

400 = n/2(2(5) + (n âˆ’ 1)d)

800 = n (10 + nd – d)

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

âŸ¹ nÂ = 16

Putting n in (1), we find d

nd – d = 40

16d – d = 40

15d = 40

d =Â 8/3

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

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

**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}Â = 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 of 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, n ^{th}Â term is -11 and the sum of first n term is 66. Find n and the d, the common difference.Â **

**Solution:**

Given,

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

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

And, 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 + (âˆ’11)]

66 = (n/2)[22 âˆ’ 11]

(66)(2) = n(11)

6 Ã— 2 = n

n = 12

Now, for finding d

We know that, l = a + (n – 1)d

– 11 = 22 + (12 – 1)d

-11 = 22 + 11d

11d = â€“ 33

d = â€“ 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 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}Â = a + (n – 1)d

âŸ¹ 49 = 7 + (n – 1)d

âŸ¹ 43 = nd – d

âŸ¹ nd – d = 42Â Â ….. (1)

Next,

S_{n}Â = n/2(2(7) + (n âˆ’ 1)d)

âŸ¹ 840 = n[14 + nd – d]

âŸ¹ 840 = n[14 + 42] [using (1)]

âŸ¹ 840 = 54n

âŸ¹ n = 15Â …. (2)

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

nd – d = 42

âŸ¹ 15d – d = 42

âŸ¹ 14d = 42

âŸ¹ 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}Â = a + (n – 1)d

âŸ¹ 45 = 5 + (n – 1)d

âŸ¹ 40 = nd – d

âŸ¹ nd – d = 40Â Â ….. (1)

Next,

S_{n}Â = n/2(2(5) + (n âˆ’ 1)d)

âŸ¹ 400 = n[10 + nd – d]

âŸ¹ 800 = n[10 + 40] [using (1)]

âŸ¹ 800 = 50n

âŸ¹ n = 16Â …. (2)

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

nd – d = 40

âŸ¹ 16d – d = 40

âŸ¹ 15d = 40

âŸ¹ d = 8/3

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

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

**Solution:**

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

And we know that, sum of first n terms is:

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

Also, nth term is given by a_{n}Â = a + (n – 1)d

From the question, we have

S_{q}Â = 162 andÂ a_{6Â }: a_{13}Â = 1 : 2

So,

2a_{6Â }= a_{13}

âŸ¹ 2 [a + (6 – 1d)] = a + (13 – 1)d

âŸ¹ 2a + 10d = a + 12d

âŸ¹ a = 2dÂ Â …. (1)

And, S_{9}Â = 162

âŸ¹Â S_{9}Â = 9/2(2a + (9 âˆ’ 1)d)

âŸ¹ 162 =Â 9/2(2a + 8d)

âŸ¹ 162 Ã— 2 = 9[4d + 8d]Â [from (1)]

âŸ¹ 324 = 9 Ã— 12d

âŸ¹ d = 3

âŸ¹ a = 2(3) [from (1)]

âŸ¹ a = 6

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

For the 15^{th} term, a_{15}Â = a + 14d = 6 + 14 Ã— 3 = 6 + 42

Therefore, a_{15}Â = 48

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

**Solution:**

Letâ€™s consider a to be the first term and d be the common difference.

And we know that, sum of first n terms is:

S_{n}Â = n/2(2a + (n âˆ’ 1)d)Â and n^{th}Â term is given by: a_{n}Â = a + (n – 1)d

Now, from the question we have

S_{10}Â = 120

âŸ¹Â 120 = 10/2(2a + (10 âˆ’ 1)d)

âŸ¹Â 120 = 5(2a + 9d)

âŸ¹ 24 = 2a + 9dÂ …. (1)

Also given that, a_{10}Â = 21

âŸ¹Â 21 = a + (10 – 1)d

âŸ¹Â 21 = a + 9dÂ …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

âŸ¹a = 3

Now, on 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}Â term is given by

a_{n}Â = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

= 2n + 1

Hence, the n^{th}Â term of the A.P is (a_{n}) = 2n + 1.

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

**Solution:**

Letâ€™s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

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

S_{7}Â = 63

And sum of next 7 terms is 161.

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

S_{14 }= 63 + 161 = 224

Now, having

S_{7}Â = 7/2(2a + (7 âˆ’ 1)d)

âŸ¹ 63(2) = 7(2a + 6d)

âŸ¹ 9 Ã— 2 = 2a + 6d

âŸ¹ 2a + 6d = 18Â . . . . (1)

And,

S_{14}Â = 14/2(2a + (14 âˆ’ 1)d)

âŸ¹ 224 = 7(2a + 13d)

âŸ¹ 32 = 2a + 13dÂ …. (2)

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

âŸ¹ 13d – 6d = 32 – 18

âŸ¹ 7d = 14

âŸ¹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from n^{th}Â term

âŸ¹ a_{28Â }= a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

Therefore, the 28^{th}Â term is 57.

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

**Solution:**

Given that,

S_{17}Â = 182

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

S_{nÂ }= n/2(2a + (n âˆ’ 1)d)

So,

S_{7}Â = 7/2(2a + (7 âˆ’ 1)d)

182 Ã— 2 = 7(2a + 6d)

364 = 14a + 42d

26 = a + 3d

a = 26 – 3dÂ … (1)

Also, itâ€™s given that 4^{th}Â term and 17^{th}Â term are in a ratio of 1: 5. So, we have

âŸ¹ 5(a_{4}) = 1(a_{17})

âŸ¹ 5 (a + 3d) = 1 (a + 16d)

âŸ¹ 5a + 15d = a + 16d

âŸ¹ 4a = dÂ …. (2)

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

âŸ¹ 4 ( 26 – 3d ) = d

âŸ¹ 104 – 12d = d

âŸ¹ 104 = 13d

âŸ¹ d = 8

Putting d in (2), we get

âŸ¹ 4a = d

âŸ¹ 4a = 8

âŸ¹ 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 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,

a_{20} = (-4(20) + 15) = -65

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

So, using the formula

S_{n}Â = n/2(a + l)

S_{20}Â = 20/2(11 + (-65))

= 10(-54)

= -540

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

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

**Solution:**

Letâ€™s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

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

S_{10}Â = -150

And the sum of next 10 terms is -550.

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

S_{20 }= -150 + -550 = -700

Now, having

S_{10}Â = 10/2(2a + (10 âˆ’ 1)d)

âŸ¹ -150 = 5(2a + 9d)

âŸ¹ -30 = 2a + 9d

âŸ¹ 2a + 9d = -30Â . . . . (1)

And,

S_{20}Â = 20/2(2a + (20 âˆ’ 1)d)

âŸ¹ -700 = 10(2a + 19d)

âŸ¹ -70 = 2a + 19dÂ …. (2)

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

âŸ¹ 19d – 9d = -70 â€“ (-30)

âŸ¹ 10d = -40

âŸ¹ 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,â€¦..

**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,

First term of the A.P is 1505 and

S_{14} = 1505

We know that, the sum of first n terms is

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

So,

S_{14} = 14/2(2(10) + (14 âˆ’ 1)d)Â = 1505

7(20 + 13d) = 1505

20 + 13d = 215

13d = 215 â€“ 20

d = 195/13

d =15

Thus, the 25^{th} term is given by

a_{25} = 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 of the A.P.**

**Solution:**

Given,

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

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

And, 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 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 â€“ 9 = 350

9n = 350 â€“ 8

n = 342/9

n = 38

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

S_{n} = n/2 (a + l)

= 38/2(17 + 350)

= 19(367)

= 6973

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

**37. Find the number of terms of the A.P. â€“12, â€“9, â€“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,

First term, aÂ = -12

Common difference, d = a_{2}Â – a_{1}Â = â€“ 9 â€“Â (- 12)

d = – 9 + 12 = 3

And, we know that n^{th}Â term =Â a_{n}Â = a + (n – 1)d

âŸ¹ 21 = -12 + (n – 1)3

âŸ¹ 21 = -12 + 3n – 3

âŸ¹ 21 = 3n – 15

âŸ¹ 36 = 3n

âŸ¹Â 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

âŸ¹ S_{12Â }+ 12Â = 12/2[a + l] + 12

= 6[-12 + 21] + 12

= 6 Ã— 9 + 12

= 66

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