Contents

**GSEB Solutions for Class 10 mathematics – Arithmetic Progression (English Medium)**

**Exercise – 5.1**

**Question 1:**

Given a and d for the following A.P., find the following A.P. :

- a = 3, d = 2
- a = -3, d = -2
- a = 100, d = -7
- a = -100, d = 7
- a = 1000, d = -100

**Question 1(1):**

**Solution :**

For the given A.P. the first term a = 3 and the common difference d = 2

∴ T_{1 }= a = 3,

T_{2 }= T_{1 }+ d = 3 + 2 = 5,

T_{3 }= T_{2 }+ d = 5 + 2 = 7,

T_{4 }= T_{3 }+ d = 7 + 2 = 9 and

T_{n}= a + (n – 1)d

= 3 + (n – 1)(2)

T_{n} = 2n + 1

Thus, the required A.P. is 3, 5, 7, 9, …..

**Question 1(2):**

**Solution :**

Here the first term a = -3 and the common difference d = -2

∴ T_{1 }= a = -3,

T_{2 }= T_{1 }+ d = -3 + (-2) = -5,

T_{3 }= T_{2 }+ d = -5 + (-2) = -7,

T_{4 }= T_{3 }+ d = -7 + (-2) = -9 and

T_{n}= a + (n – 1)d

= -3 + (n – 1) (-2)

= -3 – 2n + 2

T_{n} = -2n – 1

Thus, the required A.P is -3, -5, -7, -9, …..

**Question 1(3):**

**Solution :**

Here the first term a = 100 and the common difference d = -7

∴ T_{1 }= a = 100,

T_{2 }= T_{1 }+ d = 100 + (-7) = 93,

T_{3 }= T_{2 }+ d = 93 + (-7) = 86,

T_{4 }= T_{3 }+ d = 86 + (-7) = 79 and

T_{n}= a + (n – 1)d

= 100 + (n – 1)(-7)

T_{n} = -7n + 107

Thus, the required A.P. is 100, 93, 86, 79, …..

**Question 1(4):**

**Solution :**

Here the first term a = -100 and the common difference d = 7

∴ T_{1 }= a = -100,

T_{2 }= T_{1 }+ d = -100 + (7) = -93,

T_{3 }= T_{2 }+ d = -93 + (7) = -86,

T_{4 }= T_{3 }+ d = -86 + (7) = -79 and

T_{n}= a + (n – 1)d

= -100 + (n – 1)(7)

T_{n}= 7n – 107

Thus, the required A.P. is -100, -93, -86, -79, …..

**Question 1(5):**

**Solution :**

Here the first term a = 1000 and the common difference d = -100.

∴ T_{1 }= a = -1000,

T_{2 }= T_{1 }+ d = -1000 + (-100) = 900,

T_{3 }= T_{2 }+ d = 900 + (-100) = 800,

T_{4 }= T_{3 }+ d = 800 + (-100) = 700 and

T_{n}= a + (n – 1)d

= -1000 + (n – 1)(-100)

T_{n}= -100n + 1100

Thus, the required A.P. is 1000, 900, 800, 700, …..

**Question 2:**

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term :

- 5, -5, 5, -5,……
- 2, 2, 2, 2,……
- 1, 11, 111, 1111,….
- 5, 15, 25, 35, 45,…
- 17, 22, 27, 32,…
- 101, 99, 97, 95,…
- 201, 198, 195, 192,…
- Natural numbers which are consecutive multiples of 5 in increasing order.
- Natural numbers which are multiples of 3 or 5 in increasing order.

**Question 2(1):**

**Solution :**

For the sequence 5, -5, 5, -5, ….

T_{2 }– T_{1 }= (-5) – 5 = -10

T_{3 }– T_{2 }= 5 – (-5) = 10

But, T_{2} – T_{1 }≠ T_{3 }– T_{2}

Hence, the given sequence is not an A.P.

**Question 2(2):**

**Solution :**

For the sequence 2, 2, 2, 2, …..

T_{2 }– T_{1 }= 0

T_{3 }– T_{2 }= 0 ……

But, for an A.P, the common difference must be a non-zero constant.

Hence, the given sequence is not an A.P.

**Question 2(3):**

**Solution :**

For the sequence 1, 11, 111, 1111, ….

T_{2 }– T_{1 }= 11 – 1 = 10 and

T_{3 }– T_{2 }= 111 – 11 = 100

But, T_{2 }– T_{1 }≠ T_{3 }– T_{2}

Hence, the given sequence is not an A.P.

**Question 2(4):**

**Solution :**

For the sequence 5, 15, 25, 35, 45, …..

T_{2 }– T_{1 }= 15 – 5 = 10

T_{3 }– T_{2 }= 25 – 15 = 10

T_{4 }– T_{3 }= 35 – 25 = 10

T_{5 }– T_{4 }= 45 – 35 = 10

So,

T_{2 }– T_{1 }= T_{3 }– T_{2 }= T_{4 }– T_{3 }= T_{5 }– T_{4 … }= 10

Assuming that the pattern continues, the given sequence is an A.P.

Here, first term a = 5 and common difference d = 10.

Then,

T_{n}= a + (n – 1)d

∴ T_{n}= 5 + (n – 1)10

∴ T_{n}= 10n – 5

**Question 2(5):**

**Solution :**

For the sequence 17, 22, 27, 32, ……

T_{2 }– T_{1 }= 22 – 17 = 5

T_{3 }– T_{2 }= 27 – 22 = 5

T_{4 }– T_{3 }= 32 – 27 = 5

So, T_{2 }– T_{1 }= T_{3 }– T_{2 }= T_{4 }– T_{3 }= 5

Assuming that the pattern continues, the given sequence is an A.P.

Here the first term a = 17 and common difference d = 5.

Then, T_{n}= a + (n – 1)d

∴T_{n}= 17 + (n – 1)5

∴T_{n} = 5n + 12

**Question 2(6):**

**Solution :**

For the sequence 101, 99, 97, 95, ……

T_{2 }– T_{1 }= 99 – 101 = -2

T_{3 }– T_{2 }= 97 – 99 = -2

T_{4 }– T_{3 }= 95 – 97 = -2

So, T_{2 }– T_{1 }= T_{3 }– T_{2 }= T_{4 }– T_{3 }… _{ }= -2

Assuming that the pattern continues, the given sequence is an A.P.

Here the first term a =101 and common difference d = -2.

Now, T_{n}= a + (n – 1)d

∴ T_{n}= 101 + (n – 1)( -2)

∴ T_{n}= -2n + 103

**Question 2(7):**

**Solution :**

For the sequence 201, 198, 195, 192, …..

T_{2 }– T_{1 }= 198 – 201 = -3

T_{3 }– T_{2 }= 195 – 198 = -3

T_{4 }– T_{3 }= 192 – 195 = -3

Then,

T_{2 }– T_{1 }= T_{3 }– T_{2 }= T_{4 }– T_{3 }= -3

Assuming that the pattern continues, the given sequence is an A.P.

Here the first term a = 201 and common difference d = -3.

Now, T_{n}= a + (n – 1)d

∴ T_{n}= 201 + (n – 1)(-3)

∴ T_{n}= -3n + 204

**Question 2(8):**

**Solution :**

The given sequence is 5, 10, 15, 20, …….

Here,

T_{1 }= 10 – 5 = 5

T_{3 }– T_{2 }= 15 – 10 = 5

T_{4 }– T_{3 }= 20 – 15 = 5

So, T_{2 }– T_{1 }= T_{3 }– T_{2 }= T_{4 }– T_{3 }= 5

Assuming that the pattern continues, the given sequence is an A.P.

Here the first term a = 5 and common difference d = 5.

Then,

T_{n}= a + (n – 1)d

∴ T_{n}= 5 + (n – 1)(5)

∴ T_{n}= 5n

**Question 2(9):**

**Solution :**

The given sequence is 3, 5, 6, 9, 10, 12, 15, ……

Here, T_{1} = a = 3

T_{2} – T_{1 }= 5 – 3 = 2

T_{3 }– T_{2 }= 6 – 5 = 1

As T_{2 }– T_{1 }≠ T_{3 }– T_{2}

The given sequence is not an A.P.

**Question 3:**

Find the nth term of the following A.P.’s .

- 2, 7, 12, 17,…
- 200, 195, 190, 185,….
- 1000, 900, 800,…
- 50, 100, 150, 200,..
- \(\frac{1}{2}\), \(\frac{3}{2}\), \(\frac{5}{2}\), \(\frac{7}{2}\), \(\frac{9}{2}\) ,…..
- 1.1, 2.1, 3.1, 4.1,……
- 1.2, 2.3, 3.4, 4.5,…
- \(\frac{5}{3}\), \(\frac{7}{3}\), 3 , \(\frac{11}{3}\), \(\frac{13}{3}\), 5,………

**Solution :**

1. For the given A.P.,

First term = a = 2 and

Common difference = d = 7 – 2 = 5

Now n^{th} term,

T_{n}= a + (n – 1)d

= 2 + (n – 1)(5)

∴ T_{n}= 5n – 3

2. For the given A.P.,

First term = a = 200

Common difference = d = 195 – 200 = -5

Now n^{th} term,

T_{n}= a + (n – 1)d

= 200 + (n – 1)(-5)

∴ T_{n}= 205 – 5n

3. For the given A.P.,

First term = a = 1000

Common difference = d = 900 – 1000 = -100

Now n^{th} term,

T_{n}= a + (n – 1)d

= 1000 + (n – 1)(-100)

∴ T_{n}= 1100 – 100n

4. For the given A.P.,

First term = a = 50 Common difference = d = 100 – 50 = 50

Now n^{th} term

T_{n}_{ }= a + (n – 1)d

= 50 + (n – 1)(50)

∴ T_{n}= 50n

5.

For the given A.P.,

6. First term = a = 1.1

Common difference = d = 2.1 – 1.1= 1

Now nth term T_{n}= a + (n – 1)d

= 1.1 + (n – 1)(1)

∴ T_{n}= n + 0.1

7. For the given A.P.,

First term = a = 1.2

Common difference = d = 2.3 – 1.2 = 1.1

Now nth term

T_{n}= a + (n – 1)d

= 1.2 + (n – 1)(1.1)

∴ T_{n}= 1.1n + 0.1

8.

**Question 4:**

Find A.P. if T_{n}, T_{m} are as given below :

- T
_{7}, = 12, T_{12}= 72 - T
_{2}= 1, T_{12}= -9

**Question 4(1):**

**Solution :**

**Question 4(2):**

**Solution :**

**Question 5:**

- In an A.P., T
_{3}= 8, T_{10}= T_{6}+ 20. Find the A.P. - In an A.P. 5th term is 17 and 9th term exceeds 2nd term by 35. Find the A.P.

**Question 5(1):**

**Solution :**

**Question 5(2):**

**Solution :**

**Question 6:**

Can any term of A.P., 12, 17, 22, 27,… be zero ? Why ?

**Solution :**

**Question 7:**

Can any term of A.P., 201, 197, 193,… be 5 ? Why ?

**Solution :**

For the given A.P., 201, 197, 193. ……

a = 201 and d = 197 – 201 = -4

Assume that T_{n}= 5

5 = a + (n – 1)d

∴ 5 = 201 + (n – 1)(-4)

∴ 5 = 205 – 4n

∴ 4n = 200

∴ n = 50 ∊ N

Thus, the 50^{th} term of the given A.P. is 5.

**Question 8:**

Which term of A.P., 8, 11, 14, 17,.. is 272 ?

**Solution :**

For the given A.P. 8, 11, 14, 17, ……….

a = 8 and d = 11 – 8 = 3

Let the n^{th} term of the A.P. be 272.

T_{n}= a + (n – 1)d

∴ 272 = 8 + (n – 1)(3)

∴ 272 = 3n + 5

∴ 3n = 267

∴ n = 89

Hence the 89^{th} term of the given A.P. is 272.

**Question 9:**

Find the 10th term from end for A.P., 3, 6, 9, 12,… 300.

**Solution :**

For the given A.P., 3, 6, 9, 12, ………, 300

a = 3 and d = 6 – 3 = 3

Here T_{n} = 300, then

300 = 3 + (n – 1)(3) [∵T_{n}= a + (n – 1)d]

∴ 3(n – 1) = 297

∴ n – 1 = 99

∴ n = 100

Now, the 10^{th} term from the end is 91^{st} term of the A.P.

[∵ 91^{st} term = 100 – 10 + 1]

T_{91 }= a + (91 – 1)d

= 3 + 90 × 3

= 273

Thus, the 10^{th} term from the end of the A.P. is 273.

**Question 10:**

Find the 15th term from end for A.P., 10, 15, 20, 25, 30,…,1000.

**Solution :**

For the given A.P., 10, 15, 20, 25, 30, …….. ,1000

a = 10 and d = 15 – 10 = 5,

Here T_{n} = 1000

1000 = 10 + (n – 1)5 [∵ T_{n}= a + (n – 1)d]

∴ 990 = 5(n – 1)

∴ 198 = n – 1

∴ n = 199

Now, the 15^{th} term from the end is 185^{th} term of the A.P. (∵ 185^{th} term = 199 – 15 + 1)

T_{185 }= 10 + (185 – 1)( 5)

= 10 + 184 × 5

= 10 + 920

∴ T_{185 }= 930

Thus, the 15^{th} term from the end of the A.P. is 930.

**Question 11:**

If in an AP., T_{7} = 18, T_{18} = 7, find T_{101}.

**Solution :**

**Question 12:**

If in an A.P., T_{m} = n, T_{n} = m, prove d = -1.

**Solution :**

**Exercise – 5.2**

**Question 1:**

Find the sum of the first n terms of the A.P. as asked for :

- 2, 6, 10, 14,… upto 20 terms
- 5, 7, 9, 11,…… upto 30 terms
- -10, -12, -14, -16,… upto 15 terms
- 1, 1.5, 2, 2.5, 3,……
- \(\frac{1}{3}\), \(\frac{4}{3}\), \(\frac{7}{3}\), \(\frac{10}{3}\),….. upto 18 terms

**Question 1(1):**

**Solution :**

**Question 1(2):**

**Solution :**

**Question 1(3):**

**Solution :**

**Question 1(4):**

**Solution :**

**Question 1(5):**

**Solution :**

**Question 2:**

Find the sums indicated below :

- 3 + 6 + 9 + … + 300
- 5 + 10 + 15 + ….. + 100
- 7 + 12 + 17 + 22 + ….. + 102
- (-100) + (-92) + (-84) + …… + 92
- 25 + 21 + 17 + 13 + ….. + (-51)

**Question 2(1):**

**Solution :**

**Question 2(2):**

**Solution :**

**Question 2(3):**

**Solution :**

**Question 2(4):**

**Solution :**

**Question 2(5):**

**Solution :**

**Question 3:**

For a given A.P. with

- a = 1, d = 2, find S
_{10}. - a = 2, d = 3, find S
_{30}. - S
_{3}= 9, S_{7}= 49, find Sn and S_{10}. - T
_{10}= 41, S_{10}= 320, find T_{n}, S_{n}. - S
_{10}= 50, a = 0.5, find d. - S
_{20}= 100, d = -2, find a.

**Question 3(1):**

**Solution :**

**Question 3(2):**

**Solution :**

**Question 3(3):**

**Solution :**

**Question 3(4):**

**Solution :**

**Question 3(5):**

**Solution :**

**Question 3(6):**

**Solution :**

**Question 4:**

How many terms of A.P., 2, 7, 12, 17,… add upto 990 ?

**Solution :**

**Question 5:**

The first term of finite A.P. is 5, the last term is 45 and the sum is 500. Find the number of terms.

**Solution :**

**Question 6:**

If the first term and the last term of a finite A.P. are 5 and 95 respectively and d = 5, find n and S_{n}.

**Solution :**

**Question 7:**

The sum of first n terms of an A.P. is 5n – 2n^{2}. Find the A.P. i.e. a and d.

**Solution :**

Here S_{n}= 5n – 2n^{2}

∴ S_{n-1 }= 5(n – 1) – 2(n – 1)^{2}

= 5n – 5 – 2n^{2 }+ 4n – 2

= 9n – 2n^{2 }– 7

Now, T_{n}= S_{n}– S_{n-1}

= (5n – 2n^{2}) – (9n – 2n^{2 }– 7)

∴ T_{n}= 7 – 4n, where n > 1 …..(i)

Also, a = T_{1} = 7 – 4(1) = 3

∴ a = 3

Now,T_{2 }= 7 – 4(2) = 7 – 8 = (-1)

d = T_{2 }– T_{1 }= (-1) – 3

∴ d = -4

**Question 8:**

Find the sum of all three digit numbers divisible by 3.

**Solution :**

**Question 9:**

Find the sum of all odd numbers from 5 to 205.

**Solution :**

**Question 10:**

Which term of A.P. 121, 117, 113,….. is its first negative term? If it is the nth term, find S_{n}.

**Solution :**

**Exercise – 5**

**Question 1:**

If T_{n} = 6n + 5, find S_{n}.

**Solution :**

**Question 2:**

If S_{n} = n^{2} + 2B, find T_{n}.

**Solution :**

Here S_{n}= n^{2 }+ 2n

∴ S_{n-1}

= (n – 1)^{2 }+ 2(n – 1)

= n^{2 }– 2n + 1 + 2n – 2

= n^{2 }– 1

Now, T_{1 }= S_{1 }= (1)^{2 }+ 2(1) = 3

And, T_{n}

= S_{n}– S_{n-1}, where n > 1

= (n^{2 }+ 2n) – (n^{2 }– 1)

∴ T_{n}= 2_{n }+ 1

**Question 3:**

If the gum of firgt n terms of A.P, 30, 27, 24, 21…………. is 120, find number of terms and the last term.

**Solution :**

**Question 4:**

Which term of A.P., 100, 97, 94, 91… will be its first -ve term ?

**Solution :**

**Question 5:**

Find the sum of all 3 digit natural multiples of 6.

**Solution :**

**Question 6:**

The ratio of the sum to m terms to sum to n terms of an A.P. is \(\frac{{{m}^{2}}}{{{n}^{2}}}\). Find the ratio of its mth Term tor its nth term.

**Solution :**

**Question 7:**

Sum to first l, m, n terms of A.P. are p, q, r. Prove that \(\frac{p}{l}\) (m – n) + \(\frac{q}{m}\) (n – l) + \(\frac{r}{n}\) (l – m) = 0

**Solution :**

**Question 8:**

The ratio of sum to n terms of two A.P.’s is \(\frac{8n+1}{7n+3}\) for every n ∈ N. Find the ratio of their 7th terms and mth terms.

**Solution :**

**Question 9:**

Three numbers in A.P. have the sum 18 and the sum Of their squares is 180. Find the numbers in the increasing order.

**Solution :**

Suppose the three numbers in A.P. are

a – d, a and a + d.

According to the first condition.

(a – d) + a + (a + d) = 18

∴ 3a = 18

∴ a = 6

According to the second condition,

(a – d)^{2 }+ a^{2 }+ (a + d)^{2 }= 180

∴ a^{2 }– 2ad + d^{2 }+ a^{2 }+ a^{2 }+ 2ad + d^{2 }= 180

∴ 3a^{2 }+ 21d^{2 }= 180

∴ 108 + 2d^{2 }= 180

∴ 2d^{2 }= 72

∴ d^{2 }= 36

∴ d = 6 or d = -6

Taking a = 6 and d = 6,

First term = a – d = 6 – 6 = 0

Second term = a = 6 and

Third term = a + d = 6 + 6 = 12

Taking a = 6 and d = -6

First term = a – d = 6 – (-6) = 12

Second term = a = 6 and

Third term a + d = 6 + (-6) = 0

Thus, the required numbers are 0, 6 and 12 or 12, 6 and 0.

Arranging the numbers in the increasing order-

0, 6 and 12.

**Question 10:**

In potato race bucket is placed at the starting point. It 5 m away the from the first potato. The rest of the potatoes are placed in a straight Iine each 3 m away from the other. Each competitor starts from the bucket. Picks up the nearest potato and runs back and drops it in the bucket and continues till all potatoes are placed in the bucket. What is total distance covered if 15 potatoes

are placed in the race ?

If the distance covered is 1340 m, find the number of potatoes?

**Solution :**

**Question 11:**

A ladder has rungs 25 cm apart. The rungs decrease uniformly from 60 cm at bottom to 40 cm at top. If the distance between the top rung and the bottom rung is 2.5 m, find length of the wood required.

**Solution :**

**Question 12:**

A man purchased LCD TV for ₹ 32.500. He paid ₹ 200 initially and increasing the payment by ₹ 150 every month. How many months did he take to make the complete payment ?

**Solution :**

**Question 13:**

In an A.P., T_{1} = 22, T_{n} = -11, S_{n} = 66. find n.

**Solution :**

**Question 14:**

In an A.P. a = 8, T_{n} = 33, S_{n} = 123, find d and n.

**Solution :**

**Question 15:**

Select a proper option (a), (b), (c) or (d) from given option :

**Question 15(1):**

If T_{3} = 8, T_{7} = 24, then T_{10} = …….

**Solution :**

**Question 15(2):**

If S_{n} = 2n^{2} + 3n, then d =

**Solution :**

b. 4

Here, a = T_{1 }= S_{1} = 2(1)^{2 }+ 3(1) = 5

and a + (a + d) = S_{2} = 2(2)^{2 }+ 3(2)

∴ 2a + d = 14

∴ 2(5) + d = 14

∴ d = 4

**Question 15(3):**

If the sum of the three consecutive terms of A.P. is 48 and the product of the first and the last is 252, then d = ………

**Solution :**

a. 2

Let the three consecutive terms of the A.P. be

a – d, a, a + d.

∴ (a – d) + a + (a + d) = 48 (first condition)

∴ 3a = 48

∴ a = 16

(a – d)(a + d) = 252 (second condition)

∴ a^{2 }– d^{2 }= 252

∴ (16)^{2 }– d^{2 }= 252

∴ 256 – 252 = d^{2}

∴ d^{2 }= 4

∴ d = 2 or d = -2

But (-2) is not there in the given options.

∴ d = 2

**Question 15(4):**

If a = 2 and d = 4, then S_{20} = …..

**Solution :**

**Question 15(5):**

If 3 + 5 + 7 + 9 +….. upto n terms = 288, then n = ……

**Solution :**

c. 16

3 + 5 + 7 + 9 + ….. upto n terms = 288

∴ 1 + (3 + 5 + 7 + 9 + …. upto n terms) = 288 + 1

∴ 1 + 3 + 5 + …… upto (n + 1) terms = 289

∴ (n + 1)^{2 }= (17)^{2} [∵1 + 3 + 5+ ….. n terms = n^{2}]

∴ n + 1 = 17

∴ n = 16

**Question 15(6):**

Four numbers are in A.P. and their sum is 72 and the largest of them is twice the smallest. Then the numbers are ..

**Solution :**

b. 12, 16, 20, 24

Let the four numbers in A.P. be

a – 3d, a – d, a + d and a + 3d.

∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 72

∴ 4a = 72

∴ a = 18

Also,

a + 3d = 2(a – 3d)

∴ a + 3d = 2a – 6d

∴ 9d = a

∴ 9d = 18

∴ d = 2

∴ First number = 18 – 3(2) = 12,

Second number = 18 – 2 = 16,

Third number = 18 + 2 = 20,

Fourth number = 18 + 3(2) = 24

**Question 15(7):**

If S_{1} = 2 + 4 +…+ 2n and S_{2} = 1 + 3 +…+ (2n – 1), then S_{1} : S_{2} = ……

**Solution :**

**Question 15(8):**

For A.P., S_{n} – 2S_{n} – 1 + S_{n} – 2 = …… (n > 2)

**Solution :**

b. d

For S_{n}– 2S_{n-1 }+ S_{n-2}

= (S_{n}– S_{n-1}) – (S_{n-1 }– S_{n-2})

= T_{n}– T_{n-1}

= d

**Question 15(9):**

If S_{m} = n and S_{n} = m then S_{m + n} = …..

**Solution :**

**Question 15(10):**

If T_{4} = 7 and T_{7} = 4, then T_{10} = ……

**Solution :**

**Question 15(11):**

lf 2k + 1, 13, 5k – 3 are three consecutive terms of A.P., then k = …..

**Solution :**

c. 4

2k + 1, 13, 5k – 3 are three consecutive terms of the A.P.

∴ 13 – (2k + 1) = (5k – 3) – 13

∴ 13 + 13 = 5k – 3 + 2k + 1

∴ 26 = 7k – 2

∴ 28 = 7k

∴ k = 4

**Question 15(12):**

(1) + (1 + 1) + (1 + 1 + 1) +…+ (1 + 1 + 1 +…n – 1 times) = ……..

**Solution :**

**Question 15(13):**

In the A.P., 5, 7, 9, 11, 13, 15,… the sixth term which is prime is …….

**Solution :**

b. 19

The given A.P. is 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, ……….

Amongst them, primes are 5, 7, 11, 13, 17, 19, 23, ……. Then, the sixth prime is 19.

**Question 15(14):**

For A.P. T_{18} – T_{8} = …..

**Solution :**

**Question 15(15):**

If for A.P., T_{25} – T_{20} = 15 then d = ……

**Solution :**