An arithmetic progression is a sequence of numbers in which each term can be obtained by adding a certain quantity to its preceding term. This is an interesting topic to learn and is also one of the important chapters of ICSE Class 10 Mathematics. Students who are finding it difficult to solve problems of Selina Solutions for Class 10 Mathematics can download the PDF of solutions prepared by subject experts at BYJUâ€™S who have a vast academic experience. Here, the students can access the Selina Solutions for Class 10 Mathematics Chapter 10 Arithmetic Progression free PDF to improve their speed in solving problems as well as increase their confidence before taking the final board exam.

## Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression Download PDF

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 10 Arithmetic Progression

## Access Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression

Exercise 10(A) Page No: 137

**1. Which of the following sequences are in arithmetic progression?**

**(i)Â 2, 6, 10, 14, â€¦.**

**(ii)Â 15, 12, 9, 6, â€¦.**

**(iii)Â 5, 9, 12, 18, â€¦.**

**(iv) 1/2, 1/3, 1/4, 1/5, â€¦.**

**Solution: **

(i)Â 2, 6, 10, 14, â€¦.

Finding the difference between the terms,

d_{1} = 6 â€“ 2 = 4

d_{2} = 10 – 6 = 4

d_{3 }= 14 â€“ 10 = 4

As d_{1} = d_{2} = d_{3}, the given sequence is in arithmetic progression.

(ii)Â 15, 12, 9, 6, â€¦.

Finding the difference between the terms,

d_{1} = 12 â€“ 15 = -3

d_{2} = 9 – 12 = -3

d_{3 }= 6 â€“ 9 = -3

As d_{1} = d_{2} = d_{3}, the given sequence is in arithmetic progression.

(iii)Â 5, 9, 12, 18, â€¦.

Finding the difference between the terms,

d_{1} = 9 â€“ 5 = 4

d_{2} = 12 – 9 = 3

d_{3 }= 18 â€“ 12 = 6

As d_{1} â‰ d_{2} â‰ d_{3}, the given sequence is not in arithmetic progression.

(iv) 1/2, 1/3, 1/4, 1/5, â€¦.

Finding the difference between the terms,

d_{1} = 1/3 â€“ 1/2 = -1/6

d_{2} = 1/4 â€“ 1/3 = -1/12

d_{3 }= 1/5 â€“ 1/4 = -1/20

As d_{1} â‰ d_{2} â‰ d_{3}, the given sequence is not in arithmetic progression.

**2. The n ^{th}Â term of sequence is (2n – 3), find its fifteenth term.**

**Solution: **

Given, n^{th}Â term of sequence is (2n – 3)

So, the 15^{th} term is when n = 15

t_{15 }= 2(15) â€“ 3 = 30 â€“ 3 = 27

Thus, the 15^{th} term of the sequence is 27.

**3. If theÂ p ^{th}Â term of an A.P. is (2p + 3); find the A.P.**

**Solution: **

Given, p^{th}Â term of an A.P. = (2p + 3)

So, on putting p = 1, 2, 3, â€¦, we have

t_{1} = 2(1) + 3 = 5

t_{2} = 2(2) + 3 = 7

t_{3} = 2(3) + 3 = 9

â€¦..

Hence, the sequence A.P. is 5, 7, 9, â€¦

**4. Find the 24 ^{th}Â term of the sequence:**

**12, 10, 8,Â 6,â€¦â€¦**

**Solution:**

Given sequence,

12, 10, 8,Â 6,â€¦â€¦

The common difference:

10 â€“ 12 = -2

8 â€“ 10 = -2

6 â€“ 8 = -2 â€¦.

So, the common difference(d) of the sequence is -2 and a = 12.

Now, the general term of this A.P. is given by

t_{n} = a + (n – 1)d = 12 + (n – 1)(-2) = 12 â€“ 2n + 2 = 14 â€“ 2n

For 24^{th} term, n = 24

t_{n} = 14 â€“ 2(24) = 14 â€“ 48 = -34

therefore, the 24^{th} term is -34

**5. Find the 30 ^{th}Â term of the sequence:**

**1/2, 1, 3/2, â€¦â€¦. **

**Solution:**

Given sequence,

1/2, 1, 3/2, â€¦â€¦.

So,

a = Â½

d = 1 â€“ Â½ = Â½

We know that,

t_{n} = a + (n – 1)d

Hence, the 30^{th} term will be

t_{30} = Â½ + (30 – 1)(1/2) = Â½ + 29/2 = 30/2 = 15

**6. Find the 100 ^{th} term of the sequence:**

**âˆš3, 2âˆš3, 3âˆš3, â€¦.**

**Solution: **

Given A.P., âˆš3, 2âˆš3, 3âˆš3, â€¦.

So,

a = âˆš3

d = 2âˆš3 – âˆš3 = âˆš3

the general term is given by,

t_{n} = a + (n – 1)d

For 100^{th} term

t_{100} = âˆš3 + (100 – 1) âˆš3 = âˆš3 + 99âˆš3 = 100âˆš3

Therefore, the 100^{th} term of the given A.P. is 100âˆš3.

**7. Find the 50 ^{th} term of the sequence: **

**1/n, (n+1)/n, (2n+1)/n, â€¦â€¦**

**Solution: **

Given sequence,

1/n, (n+1)/n, (2n+1)/n, â€¦â€¦

So,

a = 1/n

d = (n+1)/n â€“ 1/n = (n+1-1)/n = 1

Then, the general term is given by

t_{n} = a + (n – 1)d

For 50^{th} term, n = 50

t_{50} = 1/n + (50 – 1)1 = 1/n + 49 = (49n + 1)/n

Hence, the 50^{th} term of the given sequence is (49n + 1)/n.

**8. Is 402Â a term of the sequence: 8, 13, 18, 23,â€¦â€¦â€¦â€¦.?**

**Solution: **

Give sequence, 8, 13, 18, 23,â€¦â€¦â€¦â€¦.

d = 13 â€“ 8 = 5 and a = 8

So, the general term is given by

t_{n} = a + (n – 1)d

t_{n} = 8 + (n – 1)5 = 8 + 5n â€“ 5 = 3 + 5n

Now,

If 402 is a term of the sequence whose n^{th} is given (3 + 5n) then n must a non-negative integer.

3 + 5n = 402

5n = 399

n = 399/5

So, clearly n is a fraction.

Thus, we can conclude that 402 is not a term of the given sequence.

**9. Find the common difference and 99 ^{th} term of term of the arithmetic progression: **

**Solution:**

Given, A.P,

i.e., 31/4, 19/2, 45/4, â€¦â€¦.

So,

a = 31/4

Common difference, d = 19/2 â€“ 31/4 = (38 – 31)/ 4 = 7/4

Then the general term of the A.P

t_{n} = a + (n – 1)d

t_{n} = 31/4 + (n – 1)(7/4) = 31/4 + 7n/4 â€“ 7/4 = 7n/4 + 31/4 â€“ 7/4 = 7n/4 + (31 â€“ 7)/4 = 7n/4 + (24)/4

t_{n} = 7n/4 + 6

Hence, the 99^{th} term is 7(99)/4 + 6 = 693/4 + 6 = 717/4 =

**10. How many terms are there in the series :**

**(i)Â 4, 7, 10, 13, â€¦â€¦â€¦â€¦,Â 148?**

**(ii)Â 0.5, 0.53, 0.56, â€¦â€¦â€¦â€¦â€¦,Â 1.1?**

**(iii) 3/4, 1, 1 1/4, â€¦â€¦., 3?**

**Solution: **

(i) Given series, 4, 7, 10, 13, â€¦â€¦â€¦â€¦,Â 148

Here,

a = 4 and d = 7 â€“ 4 = 3

So, the given term is given by t_{n} = 4 + (n – 1)3 = 4 + 3n â€“ 3

t_{n} = 1 + 3n

Now,

148 = 1 + 3n

147 = 3n

n = 147/3 = 49

Thus, there are 49 terms in the series.

(ii) Given series, 0.5, 0.53, 0.56, â€¦â€¦â€¦â€¦â€¦,Â 1.1

Here,

a = 0.5 and d = 0.53 â€“ 0.5 = 0.03

So, the given term is given by t_{n} = 0.5 + (n – 1)0.03 = 0.5 + 0.03n â€“ 0.03

t_{n} = 0.47 + 0.03n

Now,

1.1 = 0.47 + 0.03n

1.1 â€“ 0.47 = 0.03n

n = 0.63/0.03 = 21

Thus, there are 21 terms in the series.

(iii) Given series, 3/4, 1, 1 1/4, â€¦â€¦., 3

Here,

a = 3/4 and d = 1 â€“ 3/4 = 1/4

So, the given term is given by t_{n} = 3/4 + (n – 1)1/4 = (3 + n â€“ 1)/4

t_{n} = (2 + n)/ 4

Now,

3 = (2 + n)/ 4

12 = 2 + n

n = 10

Thus, there are 10 terms in the series.

Exercise 10(B) Page No: 140

**1.** **In an A.P., ten times of its tenth term is equal to thirty times of its 30 ^{th}Â term. Find its 40^{th}Â term.**

**Solution:**

Given condition,

10 t_{10 }= 30 t_{30 }in an A.P.

To find: t_{40} = ?

We know that,

t_{n }= a + (n – 1)d

So,

10 t_{10 }= 30 t_{30}

10(a + (10 – 1)d) = 30(a + (30 – 1)d)

10(a + 9d) = 30(a + 29d)

a + 9d = 3(a + 29d)

a + 9d = 3a + 87d

2a + 78d = 0

2(a + 39d) = 0

a + 39d = a + (40 – 1)d = t_{40} = 0

Therefore, the 40^{th} term of the A.P. is 0

**2. How many two-digit numbers are divisible by 3?**

**Solution:**

The 2-digit numbers divisible by 3 are as follows:

12, 15, 18, 21, â€¦â€¦â€¦, 99

Itâ€™s seen that the above forms an A.P. with

a = 12, d = 3 and last term(n^{th} term) = 99

We know that,

t_{n }= a + (n – 1)d

So,

99 = 12 + (n – 1)3

99 = 12 + 3n â€“ 3

99 = 9 + 3n

3n = 90

n = 90/3 = 30

Hence, the number of 2-digit numbers divisible by 3 is 30

**3. Which term of A.P. 5, 15, 25 â€¦â€¦â€¦â€¦ will be 130 more than its 31 ^{st}Â term?**

**Solution:**

Given A.P. 5, 15, 25, â€¦â€¦

a = 5, d = 10

From the question, we have

t_{n} = t_{31} + 130

We know that,

t_{n }= a + (n – 1)d

So,

5 + (n – 1)10 = 5 + (31 – 1)10 + 130

10n â€“ 10 = 300 + 130

10n = 430 + 10 = 440

n = 440/10 = 44

Thus, the 44^{th} term of the given A.P. is 130 more than its 31^{st} term.

**4.** **Find the valueÂ of p, if x, 2x + p and 3x + 6 are in A.P**

**Solution:**

Given that,

x, 2x + p and 3x + 6 are in A.P

So, the common difference between the terms must be the same.

Hence,

2x + p â€“ x = 3x + 6 â€“ (2x + p)

x + p = x + 6 â€“ p

2p = 6

p = 3

**5. If the 3 ^{rd}Â and the 9^{th}Â terms of an arithmetic progression are 4 andÂ -8 respectively, which term of itÂ is zero?**

**Solution:**

Given in an A.P.

t_{3} = 4 and t_{9} = -8

We know that,

t_{n }= a + (n – 1)d

So,

a + (3 – 1)d = 4 and a + (9 – 1)d = -8

a + 2d = 4 and a + 8d = -8

Subtracting both the equations we get,

6d = -12

d = -2

Using the value of d,

a + 2(-2) = 4

a â€“ 4 = 4

a = 8

Now,

t_{n} = 0

8 + (n – 1)(-2) = 0

8 â€“ 2n + 2 = 0

6 = 2n

n = 3

Thus, the 3^{rd} term of the A.P. is zero.

**6. How many three-digit numbers are divisible by 87?**

**Solution:**

The 3-digit numbers divisible by 87 are starting from:

174, 261, â€¦â€¦. , 957

This forms an A.P. with a = 174 and d = 87

And we know that,

t_{n }= a + (n – 1)d

So,

957 = 174 + (n – 1)87

783 = (n – 1)87

(n – 1) = 9

n = 10

Thus, 10 three-digit numbers are divisible by 87.

**7. For what value of n, the n ^{th}Â term of A.P 63, 65,Â 67, â€¦â€¦..Â andÂ n^{th}Â term of A.P. 3, 10, 17,â€¦â€¦..Â areÂ equal to each other?**

**Solution:**

Given,

A.P._{1} = 63, 65,Â 67, â€¦â€¦..

a = 63, d = 2 and t_{n} = 63 + (n – 1)2

A.P._{2} = 3, 10, 17, â€¦â€¦..

a = 3, d = 7 and t_{n} = 3 + (n – 1)7

Then according to the question,

n^{th} of A.P._{1} = n^{th} of A.P._{2}

63 + (n – 1)2 = 3 + (n – 1)7

60 + 2n â€“ 2 = 7n â€“ 7

58 + 7 = 5n

n = 65/5 = 13

Therefore, the 13^{th} term of both the A.P.s is equal to each other.

**8. Determine the A.P. whose 3 ^{rd}Â term is 16 and the 7^{th}Â term exceeds the 5^{th}Â term by 12.**

**Solution:**

Given,

t_{3} of an A.P. = 16 and

t_{7} = t_{5} + 12

We know that,

t_{n }= a + (n – 1)d

So,

t_{3} = a + (3 – 1)d = 16

a + 2d = 16 â€¦â€¦ (i)

And,

a + (7 – 1)d = a + (5 – 1)d + 12

6d = 4d + 12

2d = 12

d = 6

Using â€˜dâ€™ in (i) we get,

a + 2(6) = 16

a + 12 = 16

a = 4

Hence, after finding the first term = 4 and common difference = 6 the A.P. can be formed.

i.e. 4, 10, 16, 22, 28, â€¦â€¦.

Exercise 10(C) Page No: 143

**1. Find the sum of the first 22 terms of the A.P.: 8, 3, -2, â€¦â€¦â€¦..**

**Solution: **

Given, A.P. 8, 3, -2, â€¦â€¦â€¦..

Here,

a = 8

d = 3 â€“ 8 = -5

And we know that,

S_{n} = n/2[2a + (n – 1)d]

So,

S_{22} = 22/2[2(8) + (22 – 1)(-5)]

= 11[16 + (21 x -5)]

= 11[16 – 105]

= 11[-89]

= – 979

**2. How many terms of the A.P. :**

**24, 21,Â 18, â€¦â€¦â€¦Â must be taken so that their sum is 78?**

**Solution: **

Given, A.P. 24, 21,Â 18, â€¦â€¦â€¦

Here,

a = 24

d = 21 â€“ 24 = -3 and

S_{n} = 78

We know that,

S_{n} = n/2[2a + (n – 1)d]

So,

78 = n/2[2(24) + (n – 1)(-3)]

78 = n/2[48 â€“ 3n + 3]

156 = n[51 â€“ 3n]

3n^{2} â€“ 51n + 156 = 0

n^{2} â€“ 17n + 52 = 0

n^{2} â€“ 13n â€“ 4n + 52 = 0

n(n – 13) â€“ 4(n – 13) = 0

(n – 4) (n â€“ 13) = 0

So, n – 4 = 0 or n â€“ 13 = 0

Thus, n = 4 or 13

Hence, the required number of terms to be taken can be first 4 or first 13.

**3. Find the sum of 28 terms of an A.P. whose n ^{th}Â term is 8n – 5.**

**Solution: **

Given,

n^{th} term of the A.P. = 8n – 5

So,

a = 8(1) â€“ 5 = 8 â€“ 5 = 3

Here, the last term is the 28^{th} term

l = 8(28) – 5 = 224 â€“ 5 = 219

We know that,

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

Thus,

S_{28} = 28/2(3 + 219) = 14(222) = 3108

**4. (i) Find the sum of all odd natural numbers less than 50**

**Solution: **

Odd natural numbers less than 50 are:

1, 3, 5, 7, â€¦â€¦.. ,49

This forms an A.P. with a = 1, d = 2 and l = 49

And,

49 = 1 + (n – 1)2

49 = 1 + 2n â€“ 2

50 = 2n

n = 25

We know that,

S_{25} = n/2 [a + l]

= 25/2 [1 + 49]

= 25/2 [50] = 25 x 25

Thus, S_{25} = 625

**(ii) Find the sum of first 12 natural numbers each of which is a multiple of 7.**

**Solution: **

The first 12 natural numbers which are multiples of 7 are:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 and 84

This is an A.P where,

a = 7, d = 7, l = 84 and n = 12

So,

S_{12} = 12/2 [7 + 84]

= 6 [91]

= 546

**5. Find the sum of first 51 terms of an A.P. whose 2 ^{nd}Â and 3^{rd}Â terms are 14 and 18 respectively.**

**Solution: **

Given,

Number of terms of an A.P. (n) = 51

t_{2} = 14 and t_{3} = 18

So, the common difference (d) = t_{3} â€“ t_{2} = 18 â€“ 14 = 4

And,

t_{2} = a + d

14 = a + 4

a = 10

Now, we know that

S_{n} = n/2[2a + (n – 1)d]

Thus,

S_{51} = 51/2[2(10) + (51 – 1)(4)]

= 51/2[20 + 50×4]

= 51/2[20 + 200]

= 51/2[220]

= 51 x 110

= 5610

**6. The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.**

**Solution: **

Given,

S_{7 }= 49

S_{17} = 289

We know that,

S_{n} = n/2[2a + (n – 1)d]

So,

S_{7 }= 7/2[2a + (7 – 1)d] = 49

7[2a + 6d] = 98

2a + 6d = 14

a + 3d = 7 â€¦..(1)

And,

S_{17} = 17/2[2a + (17 – 1)d] = 289

17/2[2a + 16d] = 289

17[a + 8d] = 289

a + 8d = 289/17

a + 8d = 17 â€¦â€¦ (2)

Subtracting (1) from (2), we have

5d = 10

d = 2

Using value of d in (1), we get

a + 3(2) = 7

a = 7 â€“ 6 = 1

Therefore,

S_{n} = n/2[2(1) + (n – 1)(2)]

** = **n/2[2 + 2n – 2]

= n/2[2n]

S_{n }= n^{2}

**7. The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.**

**Solution: **

Given,

First term of an A.P. = 5 = a

Last term of the A.P = 45 = l

S_{n} = 1000

We know that,

S_{n} = n/2[a + l]

1000 = n/2[5 + 45]

1000 = n/2[50]

20 = n/2

n = 40

Now,

l = a + (n – 1)d

45 = 5 + (40 – 1)d

40 = 39d

d = 40/39

Therefore, the number of terms is 40 and the common difference is 40/39.

Exercise 10(D) Page No: 146

**1. Find three numbers in A.P. whose sum is 24 and whose product is 440.**

**Solution:**

Letâ€™s assume the terms in the A.P to be (a – d), a, (a + d) with common difference as d.

Given conditions,

S_{n} = 24

(a – d) + a + (a + d) = 3a = 24

a = 24/3 = 8

And,

Product of terms = 440

(a – d) x a x (a + d) = 440

a(a^{2} â€“ d^{2}) = 440

8(64 â€“ d^{2}) = 440

(64 â€“ d^{2}) = 55

d^{2} = 9

d = Â± 3

Hence,

When a = 8 and d = 3, we have

A.P. = 5, 8, 11

And, when a = 8 and d = -3 we have

A.P. = 11, 8, 5

**2. The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.**

**Solution:**

Let the consecutive terms of an A.P. be taken as (a – 1), a, (a + 1)

Given conditions,

Sum of the three consecutive terms = 21

(a – d) + a + (a + d) = 21

3a = 21

a = 7

And,

Sum of squares = (a – d)^{2} + a^{2} + (a + d)^{2} = 165

a^{2} â€“ 2ad + d^{2} + a^{2} + a^{2 }+ 2ad + d^{2} = 165

3a^{2} + 2d^{2} = 165

3(49) + 2d^{2} = 165

2d^{2} = 165 â€“ 147 = 18

d^{2} = 9

d = Â± 3

Hence,

When a = 7 and d = 3, we have

A.P. = 4, 7, 10

And, when a = 7 and d = -3 we have

A.P. = 10, 7, 4

**3. The angles of a quadrilateral are in A.P. with common difference 20 ^{o}. Find its angles.**

**Solution:**

Given, the angles of a quadrilateral are in A.P. with common difference 20^{o}.

So, let the angles be taken as x, x + 20^{o}, x + 40^{o} and x + 60^{o}.

We know that,

Sum of all the interior angles of a quadrilateral is 360^{o}

x + x + 20^{o} + x + 40^{o} + x + 60^{o}Â = 360^{o}

4x + 120^{o} = 360^{o}

4x = 360^{o} â€“ 120^{o} = 240^{o}

x = 240^{o}/ 4 = 60^{o}

Hence, the angles are

60^{o}, (60^{o} + 20^{o}), (60^{o} + 40^{o}) and (60^{o} + 60^{o})

i.e. 60^{o}, 80^{o}, 100^{o} and 120^{o}

**4. Divide 96 into four parts which are in A.P and the ratio between product of their means to product of their extremes isÂ 15: 7.**

**Solution:**

Let 96 be divided into 4 parts as (a â€“ 3d), (a – d), (a + d) and (a + 3d) which are in A.P with common difference of 2d.

Given,

(a â€“ 3d) + (a – d) + (a + d) + (a + 3d) = 4a = 96

So, a = 24

And,

(a – d)(a + d)/ (a â€“ 3d) (a + 3d) = 15/7

[a^{2}– d

^{2}]/ [a

^{2}â€“ 9d

^{2}] = 15/7

7[a^{2} – d^{2}] = 15[a^{2} â€“ 9d^{2}]

7a^{2} â€“ 7d^{2} = 15a^{2} â€“ 135d^{2}

128d^{2} = 8a^{2}

16d^{2} = a^{2} = (24)^{2}

16d^{2} = 576

d^{2} = 36

d = Â± 6

Hence,

When a = 24 and d = 6

The parts are 6, 18, 30 and 42

And, when a = 24 and d = -6

The parts are 42, 30, 18 and 6

**5. Find five numbers in A.P. whose sum is 12.5 and the ratio of the first to the last terms is 2: 3.**

**Solution:**

Let the five numbers in A.P be taken as (a â€“ 2d), (a – d), a, (a + d) and (a + 2d).

Given conditions,

(a â€“ 2d) + (a – d) + a + (a + d) + (a + 2d) = 12.5

5a = 12.5

a = 12.5/5 = 2.5

And,

(a â€“ 2d)/ (a + 2d) = 2/3

3(a â€“ 2d) = 2(a + 2d)

3a â€“ 6d = 2a + 4d

a = 10d

2.5 = 10d

d = 0.25

Therefore, the five numbers in A.P are (2.5 â€“ 2(0.25)), (2.5 â€“ 0.25), 2.5, (2.5 + 0.25) and (2.5 + 2(0.25))

i.e. 2, 2.25, 2.5, 2.75 and 3.

Exercise 10(E) Page No: 147

**1. Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h ^{-1}. The second car goes at a speed of 8 km h^{-1}Â in the first hour and thereafter increasing the speed by 0.5 km h^{-1}Â each succeeding hour. After how many hours will the two cars meet?**

Letâ€™s assume the two cars meet after n hours.

Then, this means that two cars travel the same distance in n hours.

So,

Distance travelled by the 1^{st} car in n hours = 10 x n km

Distance travelled by the 2^{nd} car in n hours = n/2[2×8 + (n â€“ 1) x 0.5)] km

10n = n/2[2×8 + (n â€“ 1) x 0.5)]

20 = [16 + 0.5n â€“ 0.5]

20 = 15.5 + 0.5n

4.5 = 0.5n

n = 9

Hence, the two cars will meet after 9 hours.

**2.** **A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.**

From the question, itâ€™s understood that

n = 7

d = -20

S_{7} = 700

We know that,

S_{n} = n/2[2a + (n – 1)d]

700 = 7/2[2a + (7 -1)(-20)]

200 = [2a + (7 -1)(-20)]

200 = 2a â€“ 120

2a = 320

a = 160

Hence, the value of each prize will be

1^{st} – 160, 2^{nd} – 140, 3^{rd} – 120, 4^{th} – 100, 5^{th} – 80, 6^{th} – 60 and 7^{th} – 40

Exercise 10(F) Page No: 147

**1. The 6 ^{th}Â term of an A.P. is 16 and the 14^{th}Â term is 32. Determine the 36^{th}Â term.**

**Solution:**

Given,

t_{6} = 16 and t_{14 }= 32

Letâ€™s take ‘a’ to be the first termÂ and ‘d’Â to be the common difference of the given A.P.

We know that,

t_{n} = a + (n – 1)d

â‡’Â a + 5d = 16Â â€¦.(1)

And,

â‡’Â a + 13d = 32Â â€¦.(2)

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

8d = 16

d = 2

Using d in (1) we get,

a + 5(2) = 16

â‡’Â a = 6

Therefore, the 36^{th}Â term = t_{36}Â = a + 35d = 6 + 35(2) = 76

**2. If the third and the 9 ^{th}Â term of an A.P. be 4 and -8 respectively, find which term is zero?**

**Solution:**

Given,

t_{3} = 4 and t_{9} = -8

We know that,

t_{n} = a + (n – 1)d

So,

4 = a + (3 – 1)d

a + 2d = 4 â€¦.. (1)

And,

-8 = a + (9 – 1)d

a + 8d = -8 â€¦.. (2)

Subtracting (1) from (2), we have

6d = -8 â€“ 4

6d = -12

d = -2

Using d in (1), we get

a + 2(-2) = 4

a = 4 + 4 = 8

Now,

t_{n} = 0 = 8 + (n – 1)(-2)

0 = 8 – 2n + 2

2n = 10

n = 5

Therefore, the 5^{th} term of the A.P. is zero.

**Â **

**3. An A.P. consists of 50 terms of which 3 ^{rd}Â term is 12 and the last term is 106. Find the 29^{th}Â term of the A.P.**

**Solution: **

Given,

Number of terms in the A.P, n = 50

And, t_{3}Â = 12

We know that,

t_{n} = a + (n – 1)d

â‡’Â a + 2d = 12Â â€¦.(1)

Last term, l = 106

t_{50}Â = 106

a + 49d = 106Â â€¦.(2)

Subtracting (1) from (2), we get

47d = 94

d = 2

a + 2(2) = 12

a = 8

Therefore, the 29^{th} term is

t_{29}Â = a + 28d = 8 + 28(2) = 8 + 56 = 64

**4. Find the arithmetic mean of:**

**(i)Â -5 and 41**

**(ii)Â 3x – 2y and 3x + 2y**

**(iii)Â (m + n) ^{2Â }and (m – n)^{2}Â **

**Solution: **

(i) Arithmetic mean of -5 and 41 = (-5 + 41)/ 2 = 36/3 = 18

(ii) Arithmetic mean of (3x â€“ 3y) and (3x + 2y) = [(3x â€“ 3y) + (3x + 2y)]/ 2 = 6x/ 2 = 3x

(iii) Arithmetic mean of (m + n)^{2} and (m – n)^{2}**Â **

**5. Find the sum of first 10 terms of the A.P.**

**4 + 6 + 8 + â€¦â€¦**

**Solution: **

Given A.P. 4 + 6 + 8 + â€¦â€¦

Here,

a = 4 and d = 6 – 4 = 2

And, n = 10

S_{10} = 10/2 [2a + (10 – 1)d]

= 5 [2(4) + 9(2)]

= 5 [8 + 18]

= 5 x 26

= 130

**6. Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.**

**Solution:**

Given,

First term, a = 3 and last term, l = 57

And, n = 20

S = n/2 (a + l)

= 20/2 (3 + 57)

= 10 (60)

= 600

**7. How many terms of the series 18 + 15 + 12 +â€¦â€¦..Â whenÂ added together will give 45?**

**Solution: **

Given series, 18 + 15 + 12 +â€¦â€¦..

Here,

a = 18 and d = 15 â€“ 18 = -3

Letâ€™s consider the number of terms to be added as ‘n’.

So, we have

S_{n} = n/2 [2a + (n – 1)d]

45 = n/2 [2(18) + (n – 1)(-3)]

90 = n[36 – 3n + 3]

90 = n[39 – 3n]

90 = 3n[13 – n]

30 = 13n – n^{2}

n^{2}Â – 13n + 30 = 0

n^{2}Â – 10n – 3n + 30 = 0

n(n – 10) – 3(n – 10) = 0

(n – 10)(n – 3) = 0

n – 10 = 0 or n – 3 = 0

n = 10 or n = 3

Therefore, the required number of terms to be added is 3 or 10.

**8. The n ^{th}Â term of a sequence is 8 – 5n. Show that the sequence is an A.P.**

**Solution:**

Given, t_{n}Â = 8 – 5n

Now, replacing n by (n + 1), we get

t_{n+1}Â = 8 – 5(n + 1) = 8 – 5n – 5 = 3 – 5n

Now,

t_{n+1}Â –Â t_{n}Â = (3 – 5n) – (8 – 5n) = -5

As, (t_{n+1}Â –Â t_{n}) is independent of n and is thus a constant.

Therefore, the given sequence having n^{th} term (8 – 5n) is an A.P.

**9. Find the general term (n ^{th}Â term) and 23^{rd}Â term of the sequence 3, 1, -1, -3, â€¦..Â .**

**Solution: **

Given sequence is 3, 1, -1, -3, â€¦..

Now,

1 – 3 = -1 – 1 = -3 – (-1) = -2

Thus, the given sequence is an A.P. where a = 3 and d = -2.

So, the general term of the A.P is given by

t_{n}Â = a + (n – 1)d

= 3 + (n – 1)(-2)

= 3 – 2n + 2

= 5 – 2n

Therefore, the 23^{rd}Â term = t_{23}Â = 5 – 2(23) = 5 – 46 = -41

**10. Which term of the sequence 3, 8, 13, ……..Â isÂ 78?**

**Solution: **

The given sequence is 3, 8,Â 13, â€¦..

Now,

8 – 3 = 13 – 8 = 5

Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.

Let the n^{th}Â term of the given A.P. be 78.

78 = 3 + (n – 1)(5)

75 = 5n – 5

5n = 80

n = 16

Thus, the 16^{th}Â term of the given sequence is 78.

**11. Is -150 a term of 11, 8, 5, 2, …….Â ?**

**Solution:**

Given sequence is 11, 8, 5,Â 2, â€¦..

Itâ€™s seen that,

8 – 11 = 5 – 8 = 2 – 5 = -3

Thus, the given sequence is an A.P. with a = 11 and d = -3.

So, the general term of the A.P. is given by

t_{n}Â = a + (n – 1)d

-150 = 11 + (n – 1)(-5)

-161 = -5n + 5

5n = 166

n = 166/5 (which is a fraction)

As the number of terms cannot be a fraction.

Hence, clearly -150 is not a term of the given sequence.

**12. How many two digit numbers are divisible by 3?**

**Solution:**

The two-digit numbers divisible by 3 are given below:

12, 15, 18, 21, â€¦.., 99

Itâ€™s seen that the above sequence form an A.P

Where,

a = 12, d = 3 and last term (l) = 99

And, the general term is given by

t_{n} = a + (n – 1)d

So,

99 = 12 + (n – 1)3

99 = 12 + 3n â€“ 3

99 = 9 + 3n

90 = 3n

n = 90/3 = 30

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

*The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.*