This exercise contains mixed problems based on all the important concepts of A.P. For acquiring a firm grip over solving problems of A.P, students can make use of the Selina Solutions for Class 10 Maths. The solutions are precise and are prepared in simple language to meet the needs of all students. Further, the solutions to this exercise are available in the Selina Solutions Concise Maths Class 10 Chapter 10 Arithmetic Progression Exercise 10(F), PDF in the links given below.

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

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

#### Exercise 10(F) Page No: 147

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

**Solution:**

Given,

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

Let’s take ‘a’ to be the first term and ‘d’ to be the common difference of the given A.P.

We know that,

t_{n} = a + (n – 1)d

⇒ a + 5d = 16 ….(1)

And,

⇒ a + 13d = 32 ….(2)

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

8d = 16

d = 2

Using d in (1) we get,

a + 5(2) = 16

⇒ a = 6

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

**2. If the third and the 9 ^{th} 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 ….. (1)

And,

-8 = a + (9 – 1)d

a + 8d = -8 ….. (2)

Subtracting (1) from (2), we have

6d = -8 – 4

6d = -12

d = -2

Using d in (1), we get

a + 2(-2) = 4

a = 4 + 4 = 8

Now,

t_{n} = 8 + (n – 1)(-2)

Let n^{th }term of this A.P. be 0

8 + (n – 1) (-2) = 0

8 – 2n + 2 = 0

10 – 2n = 0

2n = 10

n = 5

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

** **

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

**Solution: **

Given,

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

And, t_{3} = 12

We know that,

t_{n} = a + (n – 1)d

⇒ a + 2d = 12 ….(1)

Last term, l = 106

t_{50} = 106

a + 49d = 106 ….(2)

Subtracting (1) from (2), we get

47d = 94

d = 2

Substituting the value of d in equation (1), we get

a + 2(2) = 12

a = 8

Therefore, the 29^{th} term is

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

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

**(i) -5 and 41**

**(ii) 3x – 2y and 3x + 2y**

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

**Solution: **

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

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

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

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

**4 + 6 + 8 + ……**

**Solution: **

Given A.P. is 4 + 6 + 8 + ……

Here,

a = 4 and d = 6 – 4 = 2

And, n = 10

S_{n} = n/2 [2a + (n – 1)d]

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 +…….. when added together will give 45?**

**Solution: **

Given series, 18 + 15 + 12 +……..

Here,

a = 18 and d = 15 – 18 = -3

Let’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} – 13n + 30 = 0

n^{2} – 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} term of a sequence is 8 – 5n. Show that the sequence is an A.P.**

**Solution:**

Given, t_{n} = 8 – 5n

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

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

Now,

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

As, (t_{n+1} – 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} term) and 23^{rd} term of the sequence 3, 1, -1, -3, ….. .**

**Solution: **

Given sequence is 3, 1, -1, -3, …..

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 an A.P is given by

t_{n} = a + (n – 1)d

= 3 + (n – 1)(-2)

= 3 – 2n + 2

= 5 – 2n

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

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

**Solution: **

The given sequence is 3, 8, 13, …..

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} term of the given A.P. be 78.

78 = 3 + (n – 1)(5)

75 = 5n – 5

5n = 80

n = 16

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

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

**Solution:**

Given sequence is 11, 8, 5, 2, …..

It’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 an A.P. is given by

t_{n} = a + (n – 1)d

-150 = 11 + (n – 1)(-3)

-161 = -3n + 3

3n = 164

n = 164/3 (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, ….., 99

It’s clear that the above sequence forms 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 – 3

99 = 9 + 3n

90 = 3n

n = 90/3 = 30

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

This is very congenial for every student to learn and practice all