In Exercise 19.2, we shall discuss problems based on the general terms of an A.P., which includes finding the nth term in the given expression. The solutions of the exercise wise problems under this chapter are prepared by experienced faculty team at BYJUâ€™S after conducting vast research on each concept. The main aim of designing solutions is to help students with their board exam preparation and thereby, to secure good marks. On regular practice, students can achieve their goals. RD Sharma Class 11 Maths Solutions pdf are available here, students can make use of it by downloading from the links provided below.

## Download the pdf of RD Sharma Solutions for Class 11 Maths Exercise 19.2 Chapter 19 – Arithmetic Progressions

### Also, access other exercises of RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions

### Access answers to RD Sharma Solutions for Class 11 Maths Exercise 19.2 Chapter 19 – Arithmetic Progressions

**1. Find: (i) 10 ^{th}Â term of the A.P. 1, 4, 7, 10, …..**

**(ii) 18 ^{th}Â term of the A.P. âˆš2, 3âˆš2, 5âˆš2, â€¦**

**(iii) nth term of the A.P 13, 8, 3, -2, â€¦.**

**Solution:**

**(i) **10^{th}Â term of the A.P. 1, 4, 7, 10, …..

Arithmetic Progression (AP) whose common difference is = a_{n}Â â€“ a_{n-1}Â where n > 0

Let us consider, a = a_{1}Â = 1, a_{2}Â = 4 â€¦

So, Common difference, d = a_{2}Â â€“ a_{1}Â = 4 â€“ 1 = 3

To find the 10^{th} term of A.P, firstly find a_{n}

By using the formula,

a_{n}Â = a + (n-1) d

= 1 + (n-1) 3

= 1 + 3n â€“ 3

= 3n â€“ 2

When n = 10:

a_{10}Â = 3(10) â€“ 2

= 30 â€“ 2

= 28

Hence, 10^{th}Â term is 28.

**(ii) **18^{th}Â term of the A.P. âˆš2, 3âˆš2, 5âˆš2, â€¦

Arithmetic Progression (AP) whose common difference is = a_{n}Â â€“ a_{n-1}Â where n > 0

Let us consider, a = a_{1}Â = âˆš2, a_{2}Â = 3âˆš2 â€¦

So, Common difference, d = a_{2}Â â€“ a_{1}Â = 3âˆš2 â€“ âˆš2 = 2âˆš2

To find the 18^{th} term of A.P, firstly find a_{n}

By using the formula,

a_{n}Â = a + (n-1) d

= âˆš2 + (n – 1) 2âˆš2

= âˆš2 + 2âˆš2n – 2âˆš2

= 2âˆš2n – âˆš2

When n = 18:

a_{18}Â = 2âˆš2(18) – âˆš2

= 36âˆš2 – âˆš2

= 35âˆš2

Hence, 10^{th}Â term is 35âˆš2

**(iii) **nth term of the A.P 13, 8, 3, -2, â€¦.

Arithmetic Progression (AP) whose common difference is = a_{n}Â â€“ a_{n-1}Â where n > 0

Let us consider, a = a_{1}Â = 13, a_{2}Â = 8 â€¦

So, Common difference, d = a_{2}Â â€“ a_{1}Â = 8 â€“ 13 = -5

To find the n^{th} term of A.P, firstly find a_{n}

By using the formula,

a_{n}Â = a + (n-1) d

= 13 + (n-1) (-5)

= 13 â€“ 5n + 5

= 18 â€“ 5n

Hence, n^{th}Â term is 18 â€“ 5n

**2. In an A.P., show that a _{m+n}Â + a_{mâ€“n}Â = 2a_{m}.**

**Solution:**

We know the first term is â€˜aâ€™ and the common difference of an A.P is d.

Given:

a_{m+n}Â + a_{mâ€“n}Â = 2a_{m}

By using the formula,

a_{n}Â = a + (n â€“ 1)d

Now, let us take LHS: a_{m+n}Â + a_{m-n}Â

a_{m+n}Â + a_{m-n}Â = a + (m + n â€“ 1)d + a + (m – n â€“ 1)d

= a + md + nd â€“ d + a + md – nd â€“ d

= 2a + 2md â€“ 2d

= 2(a + md â€“ d)

= 2[a + d(m â€“ 1)] {âˆµÂ a_{n}Â = a + (n â€“ 1)d}

a_{m+n}Â + a_{m-n}Â = 2a_{m}

Hence Proved.

**3. (i) Which term of the A.P. 3, 8, 13,â€¦ is 248 ?**

**(ii) Which term of the A.P. 84, 80, 76,â€¦ is 0 ?**

**(iii) Which term of the A.P. 4, 9, 14,â€¦ is 254 ?**

**Solution:**

**(i) **Which term of the A.P. 3, 8, 13,â€¦ is 248 ?

Given A.P is 3, 8, 13,â€¦

Here, a_{1}Â = a = 3, a_{2}Â = 8

Common difference, d = a_{2}Â â€“ a_{1}Â = 8 â€“ 3 = 5

We know, a_{n}Â = a + (n â€“ 1)d

a_{n}Â = 3 + (n â€“ 1)5

= 3 + 5n â€“ 5

= 5n â€“ 2

Now, to find which term of A.P is 248

Put a_{n}Â = 248

âˆ´Â 5n â€“ 2 = 248

= 248 + 2

= 250

= 250/5

= 50

Hence, 50^{th}Â term of given A.P is 248.

**(ii) **Which term of the A.P. 84, 80, 76,â€¦ is 0 ?

Given A.P is 84, 80, 76,â€¦

Here, a_{1}Â = a = 84, a_{2}Â = 88

Common difference, d = a_{2}Â â€“ a_{1}Â = 80 â€“ 84 = -4

We know, a_{n}Â = a + (n â€“ 1)d

a_{n}Â = 84 + (n â€“ 1)-4

= 84 â€“ 4n + 4

= 88 â€“ 4n

Now, to find which term of A.P is 0

Put a_{n}Â = 0

88 â€“ 4n = 0

-4n = -88

n = 88/4

= 22

Hence, 22^{nd}Â term of given A.P is 0.

**(iii) **Which term of the A.P. 4, 9, 14,â€¦ is 254 ?

Given A.P is 4, 9, 14,â€¦

Here, a_{1}Â = a = 4, a_{2}Â = 9

Common difference, d = a_{2}Â â€“ a_{1}Â = 9 â€“ 4 = 5

We know, a_{n}Â = a + (n â€“ 1)d

a_{n}Â = 4 + (n â€“ 1)5

= 4 + 5n â€“ 5

= 5n â€“ 1

Now, to find which term of A.P is 254

Put a_{n}Â = 254

5n â€“ 1 = 254

5n = 254 + 1

5n = 255

n = 255/5

= 51

Hence, 51^{st}Â term of given A.P is 254.

**4. (i) Is 68 a term of the A.P. 7, 10, 13,â€¦?**

**(ii) Is 302 a term of the A.P. 3, 8, 13,â€¦?**

**Solution:**

**(i) **Is 68 a term of the A.P. 7, 10, 13,â€¦?

Given A.P is 7, 10, 13,â€¦

Here, a_{1}Â = a = 7, a_{2}Â = 10

Common difference, d = a_{2}Â â€“ a_{1}Â = 10 â€“ 7 = 3

We know, a_{n}Â = a + (n â€“ 1)d [where, a is first term or a_{1}Â and d is common difference and n is any natural number]

a_{n}Â = 7 + (n â€“ 1)3

= 7 + 3n â€“ 3

= 3n + 4

Now, to find whether 68 is a term of this A.P. or not

Put a_{n}Â = 68

3n + 4 = 68

3n = 68 â€“ 4

3n = 64

n = 64/3

64/3Â is not a natural number

Hence, 68 is not a term of given A.P.

**(ii) **Is 302 a term of the A.P. 3, 8, 13,â€¦?

Given A.P is 3, 8, 13,â€¦

Here, a_{1}Â = a = 3, a_{2}Â = 8

Common difference, d = a_{2}Â â€“ a_{1}Â = 8 â€“ 3 = 5

We know, a_{n}Â = a + (n â€“ 1)d

a_{n}Â = 3 + (n â€“ 1)5

= 3 + 5n â€“ 5

= 5n â€“ 2

To find whether 302 is a term of this A.P. or not

Put a_{n}Â = 302

5n â€“ 2 = 302

5n = 302 + 2

5n = 304

n = 304/5

304/5Â is not a natural number

Hence, 304 is not a term of given A.P.

**5. (i) Which term of the sequenceÂ 24, 23 Â¼, 22 Â½, 21 Â¾ is the first negative term?**

**Solution:**

Given:

AP: 24, 23 Â¼, 22 Â½, 21 Â¾, â€¦ = 24, 93/4, 45/2, 87/4, â€¦

Here, a_{1} = a = 24, a_{2} = 93/4

Common difference, d = a_{2} â€“ a_{1} = 93/4 â€“ 24

= (93 – 96)/4

= – 3/4

We know, a_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

We know, a_{n}Â = a + (n â€“ 1) d

a_{n}Â = 24 + (n – 1) (-3/4)

= 24 â€“ 3/4n + Â¾

= (96+3)/4 â€“ 3/4n

= 99/4 â€“ 3/4n

Now we need to find, first negative term.

Put a_{n} < 0

a_{n} = 99/4 â€“ 3/4n < 0

99/4 < 3/4n

3n > 99

n > 99/3

n > 33

Hence, 34^{th} term is the first negative term of given AP.

**(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, â€¦ is (a) purely real (b) purely imaginary ?**

**Solution:**

Given:

AP: 12 + 8i, 11 + 6i, 10 + 4i, â€¦

Here, a_{1}Â = a = 12 + 8i, a_{2}Â = 11 + 6i

Common difference, d = a_{2}Â â€“ a_{1}

= 11 + 6i â€“ (12 + 8i)

= 11 â€“ 12 + 6i â€“ 8i

= -1 â€“ 2i

We know, a_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

a_{n}Â = 12 + 8i + (n â€“ 1) -1 â€“ 2i

= 12 + 8i â€“ n â€“ 2ni + 1 + 2i

= 13 + 10i â€“ n â€“ 2ni

= (13 â€“ n) + (10 â€“ 2n) i

To find purely real term of this A.P., imaginary part have to be zero

10 â€“ 2n = 0

2n = 10

n = 10/2

= 5

Hence, 5^{th}Â term is purely real.

To find purely imaginary term of this A.P., real part have to be zero

âˆ´Â 13 â€“ n = 0

n = 13

Hence, 13^{th}Â term is purely imaginary.

**6. (i) How many terms are in A.P. 7, 10, 13,â€¦43?**

**Solution:**

Given:

AP: 7, 10, 13,â€¦

Here, a_{1}Â = a = 7, a_{2}Â = 10

Common difference, d = a_{2}Â â€“ a_{1}Â = 10 â€“ 7 = 3

We know, a_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

a_{n}Â = 7 + (n â€“ 1)3

= 7 + 3n â€“ 3

= 3n + 4

To find total terms of the A.P., put a_{n}Â = 43 as 43 is last term of A.P.

3n + 4 = 43

3n = 43 â€“ 4

3n = 39

n = 39/3

= 13

Hence, total 13 terms exists in the given A.P.

**(ii) How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, â€¦, 10/3 ?**

**Solution:**

Given:

AP: -1, -5/6, -2/3, -1/2, â€¦

Here, a_{1}Â = a = -1, a_{2}Â = -5/6

Common difference, d = a_{2}Â â€“ a_{1}Â

= -5/6 â€“ (-1)

= -5/6 + 1

= (-5+6)/6

= 1/6

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

a_{n}Â = -1 + (n â€“ 1) 1/6

= -1 + 1/6n â€“ 1/6

= (-6-1)/6 + 1/6n

= -7/6 + 1/6n

To find total terms of the AP,

Put a_{n} = 10/3 [Since, 10/3 is the last term of AP]

a_{n} = -7/6 + 1/6n = 10/3

1/6n = 10/3 + 7/6

1/6n = (20+7)/6

1/6n = 27/6

n = 27

Hence, total 27 terms exists in the given A.P.

**7. The first term of an A.P. is 5, the common difference is 3, and the last term is 80; find the number of terms.**

**Solution:**

Given:

First term, a = 5; last term, l = a_{n}Â = 80

Common difference, d = 3

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

a_{n}Â = 5 + (n â€“ 1)3

= 5 + 3n â€“ 3

= 3n + 2

To find total terms of the A.P., put a_{n}Â = 80 as 80 is last term of A.P.

3n + 2 = 80

3n = 80 â€“ 2

3n = 78

n = 78/3

= 26

Hence, total 26 terms exists in the given A.P.

**8. The 6 ^{th}Â and 17^{th}Â terms of an A.P. are 19 and 41 respectively. Find the 40^{th}Â term.**

**Solution:**

Given:

6^{th}Â term of an A.P is 19 and 17^{th}Â terms of an A.P. is 41

So, a_{6}Â = 19 and a_{17}Â = 41

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

When n = 6:

a_{6}Â = a + (6 â€“ 1) d

= a + 5d

Similarly,Â When n = 17:

a_{17}Â = a + (17 â€“ 1)d

= a + 16d

According to question:

a_{6}Â = 19 and a_{17}Â = 41

a + 5d = 19 â€¦â€¦â€¦â€¦â€¦â€¦ (i)

And a + 16d = 41â€¦â€¦â€¦â€¦.. (ii)

Let us subtract equation (i) from (ii) we get,

a + 16d â€“ (a + 5d) = 41 â€“ 19

a + 16d â€“ a â€“ 5d = 22

11d = 22

d = 22/11

= 2

put the value of d in equation (i):

a + 5(2) = 19

a + 10 = 19

a = 19 â€“ 10

= 9

As, a_{n}Â = a + (n â€“ 1)d

a_{40}Â = a + (40 â€“ 1)d

= a + 39d

Now put the value of a = 9 and d = 2 in a_{40} we get,

a_{40}Â = 9 + 39(2)

= 9 + 78

= 87

Hence, 40^{th}Â term of the given A.P. is 87.

**9. If 9 ^{th}Â term of an A.P. is Zero, prove that its 29^{th}Â term is double the 19^{th}Â term.**

**Solution:**

Given:

9^{th}Â term of an A.P is 0

So, a_{9}Â = 0

We need to prove: a_{29}Â = 2a_{19}

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

When n = 9:

a_{9}Â = a + (9 â€“ 1)d

= a + 8d

According to question:

a_{9}Â = 0

a + 8d = 0

a = -8d

When n = 19:

a_{19}Â = a + (19 â€“ 1)d

= a + 18d

= -8d + 18d

= 10d

When n = 29:

a_{29}Â = a + (29 â€“ 1)d

= a + 28d

= -8d + 28d [Since,Â a = -8d]

= 20d

= 2Ã—10d

a_{29}Â = 2a_{19} [Since,Â a_{19}Â = 10d]

Hence Proved.

**10. If 10 times the 10 ^{th}Â term of an A.P. is equal to 15 times the 15^{th}Â term, show that the 25^{th}Â term of the A.P. is Zero.**

**Solution:**

Given:

10 times the 10^{th}Â term of an A.P. is equal to 15 times the 15^{th}Â term

So, 10a_{10}Â = 15a_{15}

We need to prove: a_{25}Â = 0

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

When n = 10:

a_{10}Â = a + (10 â€“ 1)d

= a + 9d

When n = 15:

a_{15}Â = a + (15 â€“ 1)d

= a + 14d

When n = 25:

a_{25}Â = a + (25 â€“ 1)d

= a + 24d â€¦â€¦â€¦(i)

According to question:

10a_{10}Â = 15a_{15}

10(a + 9d) = 15(a + 14d)

10a + 90d = 15a + 210d

10a â€“ 15a + 90d â€“ 210d = 0

-5a â€“ 120d = 0

-5(a + 24d) = 0

a + 24d = 0

a_{25}Â = 0 [From (i)]

Hence Proved.

**11. The 10 ^{th}Â and 18^{th}Â term of an A.P. are 41 and 73 respectively, find 26^{th}Â term.**

**Solution:**

Given:

10^{th}Â term of an A.P is 41, and 18^{th}Â terms of an A.P. is 73

So, a_{10}Â = 41 and a_{18}Â = 73

We know, a_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is the common difference and n is any natural number]

When n = 10:

a_{10}Â = a + (10 â€“ 1)d

= a + 9d

When n = 18:

a_{18}Â = a + (18 â€“ 1)d

= a + 17d

According to question:

a_{10}Â = 41 and a_{18}Â = 73

a + 9d = 41 â€¦â€¦â€¦â€¦â€¦â€¦(i)

And a + 17d = 73â€¦â€¦â€¦â€¦..(ii)

Let us subtract equation (i) from (ii) we get,

a + 17d â€“ (a + 9d) = 73 â€“ 41

a + 17d â€“ a â€“ 9d = 32

8d = 32

d = 32/8

d = 4

Put the value of d in equation (i) we get,

a + 9(4) = 41

a + 36 = 41

a = 41 â€“ 36

a = 5

we know, a_{n}Â = a + (n â€“ 1)d

a_{26}Â = a + (26 â€“ 1)d

= a + 25d

Now put the value of a = 5 and d = 4 in a_{26}

a_{26}Â = 5 + 25(4)

= 5 + 100

= 105

Hence, 26^{th}Â term of the given A.P. is 105.

**12. In a certain A.P. the 24 ^{th}Â term is twice the 10^{th}Â term. Prove that the 72^{nd}Â term is twice the 34^{th}Â term.**

**Solution:**

Given:

24^{th}Â term is twice the 10^{th}Â term

So, a_{24}Â = 2a_{10}

We need to prove: a_{72}Â = 2a_{34}

_{n}Â = a + (n â€“ 1) d [where a is first term or a_{1}Â and d is common difference and n is any natural number]

When n = 10:

a_{10}Â = a + (10 â€“ 1)d

= a + 9d

When n = 24:

a_{24}Â = a + (24 â€“ 1)d

= a + 23d

When n = 34:

a_{34}Â = a + (34 â€“ 1)d

= a + 33d â€¦â€¦â€¦(i)

When n = 72:

a_{72}Â = a + (72 â€“ 1)d

= a + 71d

According to question:

a_{24}Â = 2a_{10}

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a â€“ 2a + 23d â€“ 18d = 0

-a + 5d = 0

a = 5d

Now, a_{72}Â = a + 71d

a_{72}Â = 5d + 71d

= 76d

= 10d + 66d

= 2(5d + 33d)

= 2(a + 33d) [since,Â a = 5d]

a_{72}Â = 2a_{34}Â (From (i))

Hence Proved.

Good