More problems on finding the terms of an A.P. when given certain conditions are included in this exercise. Students who want to learn the correct steps of solving problems of this exercise and chapter can make use of the Selina Solutions for Class 10 Maths. For solutions of these exercise problems, the Concise Selina Solutions for Class 10 Maths Chapter 10 Arithmetic Progression Exercise 10(B) PDF is available for reference in the links given below.

## Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression Exercise 10(B) Download PDF

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### Access Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression Exercise 10(B)

**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€¦.