# Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.4 | Set 1

**Question 1. Find:**

**(i) 10th tent of the AP 1, 4, 7, 10….**

**(ii) 18th term of the AP √2, 3√2, 5√2, …….**

**(iii) nth term of the AP 13, 8, 3, -2, ……….**

**(iv) 10th term of the AP -40, -15, 10, 35, ………….**

**(v) 8th term of the AP 11, 104, 91, 78, ……………**

**(vi) 11th tenor of the AP 10.0, 10.5, 11.0, 11.2, …………..**

**(vii) 9th term of the AP 3/4, 5/4, 7/4 + 9/4, ………..**

**Solution: **

(i)Given that,A.P. is 1, 4, 7, 10, ……….

First term(a) = 1

Common difference (d) = Second term – First term

= 4 – 1 = 3.

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 10th term in the A.P is 1 + (10 – 1)3

= 1 + 9×3 = 1 + 27 = 28

Hence 10th term of A.P is 28

(ii)Given that,A.P. is √2, 3√2, 5√2, …….

First term (a) = √2,

Common difference = Second term – First term

d = 3√2 – √2 = 2√2

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 18th term of A. P. = √2 + (18 – 1)2√2

= √2 + 17.2√2 = √2 (1+34) = 35√2

Hence 18th term of A.P is 35√2

(iii)Given that,A. P. is 13, 8, 3, – 2, …………

First term (a) = 13,

Common difference (d) = Second term first term

d = 8 – 13 = – 5

As we know that, to find nth term in an A.P = a + (n – 1)d

= 13 + (n – 1) – 5 = 13 – 5n + 5

Hence nth term of the A.P is an = 18 – 5n

(iv)Given that,A. P. is – 40, -15, 10, 35, ……….

First term (a) = -40,

Common difference (d) = Second term – fast term

d = -15 – (- 40) = 40 – 15 = 25

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 10th term of A. P. = -40 + (10 – 1)25

= – 40 + 9.25 = – 40 + 225 = 185

Hence 10th term of the A.P is 185

(v)Given that,Sequence is 117, 104, 91, 78, ………….

First term (a) = 117,

Common difference (d) = Second term – first term

d = 104 – 117 = –13

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 8th term = a + (8 – 1)d

= 117 + 7(-13) = 117 – 91 = 26

Hence 8th term of the A.P is 26

(vi)Given that,A. P is 10.0, 10.5, 11.0, 11.5,

First term(a) = 10.0,

Common difference (d) = Second term – first term

d = 10.5 – 10.0 = 0.5

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 11th term a

_{11}= 10.0 + (11 – 1)0.5= 10.0 + 10 x 0.5 = 10.0 + 5 =15.0

Hence 11th term of the A. P. is 15.0

(vii)Given that,A. P is 3/4, 5/4, 7/4, 9/4, …………

First term (a) = 3/4,

Common difference (d) = Second term – first term

d = 5/4 – 3/4 = 2/4

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 9th term a

_{9}= a + (9 – 1)d= 3/4 + 16/4 = 19/4

Hence 9th term of the A.P is 19/4.

**Question 2. Find:**

**(i) Which term of the AP 3, 8, 13, …. is 248?**

**(ii) Which term of the AP 84, 80, 76, … is 0?**

**(iii) Which term of the AP 4. 9, 14, …. is 254?**

**(iv) Which term of the AP 21. 42, 63, 84, … is 420?**

**(v) Which term of the AP 121, 117. 113, … is its first negative term?**

**Solution: **

(i)Given that,A.P. is 3, 8, 13, ………..

First term (a) = 3,

nth term is 248

Common difference (d) = Second term – first term

d = 8 – 3 = 5

As we know that, to find nth term in an A.P = a + (n – 1)d

248 = 3+(n – 1)5

248 = -2 + 5n

5n = 250

n =250/5 = 50

Hence 50th term in the A.P is 248.

(ii)Given that,A. P is 84, 80, 76, …………

First term (a) = 84

nth term is 0

Common difference (d) = a2 – a

d = 80 – 84 = – 4

As we know that, to find nth term in an A.P = a + (n – 1)d

0 = 84 + (n – 1) – 4

84 = +4(n – 1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

Hence 22nd term in the A.P is 0.

(iii)Given A. P 4, 9, 14, …………First term (a) = 4,

nth term is 254

Common difference (d) = a2 – a1

d = 9 – 4 = 5

As we know that, to find nth term in an A.P = a + (n – 1)d

4 + (n – 1)5 = 254

(n – 1)∙5 = 250

n – 1 = 250/5 = 50

n = 51

Hence 51th term in the A.P is 254.

(iv)Given that,A. P 21, 42, 63, 84, ………

a = 21,

nth term = 420,

d = a

_{2}– a_{1}= 42 – 21 = 21

As we know that, to find nth term in an A.P = a + (n – 1)d

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

Hence 20th term is 420.

(v)Given that,A.P is 121, 117, 113, ………..

First term (a) = 121,

nth term is negative i.e. an < 0,

Common difference (d) = 117 – 121 = – 4

As we know that, to find nth term in an A.P = a + (n – 1)d

121 + (n – 1) – 4 < 0

121 + 4 – 4n < 0

125 – 4n < 0

4n > 125

n > 125/4

n > 31.25

The integer which comes after 31.25 is 32.

Hence 32nd term in the A.P will be the first negative term.

**Question 3. **

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

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

**(iii) Is -150 a term of the A.P. 11, 8, 5, 2, … ?**

**Solutions: **

(i)Given that,A.P. 7, 10, 13,…

from given series,

a = 7 and d = a

_{2}– a_{1}= 10 – 7 = 3As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position 68 is present in given series,

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n – 3 = 68

3n + 4 = 68

3n = 64

n = 64/3, which is not a whole number.

Hence, 68 is not a term in the A.P.

(ii)Given, A.P. 3, 8, 13,…from given series, a = 3 and d = a

_{2}– a_{1}= 8 – 3 = 5As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position 302 is present in given series,

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n – 5 = 302

5n – 2 = 302

5n = 304

n = 304/5, which is not a whole number.

Hence, 302 is not a term in the A.P.

(iii)Given, A.P. 11, 8, 5, 2, …from given series,

a = 11 and d = a

_{2}– a_{1}= 8 – 11 = -3As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position -150 is present in given series,

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 – 3n + 3 = -150

3n = 150 + 14

3n = 164

n = 164/3, which is not a whole number.

Hence, -150 is not a term in the A.P.

**Question 4. How many terms are there in the A.P.?**

**(i) 7, 10, 13, ….., 43**

**(ii) -1, -5/6, -2/3, -1/2, … , 10/3**

**(iii) 7, 13, 19, …, 205**

**(iv) 18, 15½, 13, …., -47**

**Solution: **

(i)Given that,A.P. 7, 10, 13, ….., 43

where, a = 7 and d = a

_{2}– a_{1}= 10 – 7 = 3As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 43

7 + (n – 1)(3) = 43

7 + 3n – 3 = 43

3n = 43 – 4

3n = 39

n = 13

Hence, there are 13 terms in the given A.P.

(ii)Given that,A.P. -1, -5/6, -2/3, -1/2, … , 10/3

where, a = -1 and d = a

_{2}– a_{1}= -5/6 – (-1) = 1/6As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 10/3

-1 + (n – 1)(1/6) = 10/3

-1 + n/6 – 1/6 = 10/3

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

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

n = (20 + 6 + 1)

n = 27

Hence, there are 27 terms in the given A.P.

(iii)Given that,A.P. 7, 13, 19, …, 205

where, a = 7 and d = a

_{2}– a_{1}= 13 – 7 = 6,nth term is 205

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 205

7 + (n – 1)(6) = 205

7 + 6n – 6 = 205

6n = 205 – 1

n = 204/6

n = 34

Hence, there are 34 terms in the given A.P.

(iv)Given that,A.P. 18, 15½, 13, …., -47

where, a = 7 and d = 15½ – 18 = 5/2,

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 43

18 + (n – 1)(-5/2) = -47

18 – 5n/2 + 5/2 = -47

36 – 5n + 5 = -94

5n = 94 + 36 + 5

5n = 135

n = 27

Hence, there are 27 terms in the given A.P.

**Question 5. 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 that,

a = 5 and d = 3,

last term = 80

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, for the given A.P. an = 5 + (n – 1)3 = 3n + 2

=3n + 2 = 80

3n = 78

n = 78/3 = 26

Hence, there are 26 terms in the A.P.

**Question 6. The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.**

**Solution: **

Given that,

a

_{6}= 19 and a_{17}= 41As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a

_{6}= a + (6-1)d= a + 5d = 19 ——-(i)

Similarly,

a

_{17 }= a + (17 – 1)d= a + 16d = 41 ———-(ii)

Solving (i) and (ii),

(ii) – (i)

a + 16d – (a + 5d) = 41 – 19

11d = 22

d = 2

Using d in eqn(i), we get

a + 5(2) = 19

a = 19 – 10 = 9

Now, the 40th term is given by a40 = 9 + (40 – 1)2 = 9 + 78 = 87

Hence the 40th term is 87.

**Question 7. If 9th term of an A.P. is zero, prove its 29th term is double the 19th term.**

**Solution: **

Given that,

a

_{9}= 0As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ———-(i)

Now,

29th term is given by a

_{29}= a + (29 – 1)d=a

_{29}= a + 28dAnd, a

_{29}= (a + 8d) + 20d (using (i))= a

_{29}= 20d ———-(ii)Similarly, 19th term is given by a

_{19}= a + (19 – 1)d=a

_{19}= a + 18dAnd, a

_{19}= (a + 8d) + 10d (using (i))=a

_{19 }= 10d ———(iii)On comparing (ii) and (iii), we observe that

a

_{29}= 2(a_{19})

Hence, 29th term is double the 19th term.

**Question 8. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.**

**Solution:**

Given that,

10 times the 10th term of an A.P. is equal to 15 times the 15th term.

As we know that, to find nth term in an A.P = a + (n – 1)d

10(a

_{10}) = 15(a_{15})10(a + (10 – 1)d) = 15(a + (15 – 1)d)

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

10a + 90d = 15a + 210d

5a + 120d = 0

5(a + 24d) = 0

a + 24d = 0

a + (25 – 1)d = 0

a

_{25}= 0

Hence, the 25th term of the A.P. is zero.

**Question 9. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.**

**Solution: **

Given that,

A

_{10}= 41 and a_{18}= 73As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a

_{10}= a + (10 – 1)d= a + 9d = 41 ———(i)

Similarly,

a

_{18}= a + (18 – 1)d= a + 17d = 73 ——-(ii)

Solving (i) and (ii),

(ii) – (i)

a + 17d – (a + 9d) = 73 – 41

8d = 32

d = 4

Using d in (i), we get

a + 9(4) = 41

a = 41 – 36 = 5

Now, the 26th term is given by a26 = 5 + (26 – 1)4 = 5 + 100 = 105

Hence the 26th term is 105.

**Question 10. In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.**

**Solution: **

Given that,

24th term is twice the 10th term.

As we know that, to find nth term in an A.P = a + (n – 1)d

a

_{24}= 2(a10)a + (24 – 1)d = 2(a + (10 – 1)d)

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

a + 23d = 2a + 18d

a = 5d …. (1)

Now, the 72nd term can be expressed as:

a

_{72}= a + (72 – 1)d= a + 71d

= a + 5d + 66d

= a + a + 66d [using (1)]

= 2(a + 33d)

= 2(a + (34 – 1)d)

= 2(a34)

⇒ a

_{72}= 2(a_{34})

Hence, the 72nd term is twice the 34th term of the given A.P.

**Question 11. The 26th, 11th and the last term of an A.P. are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.**

**Solution: **

Given that,

a

_{26}= 0, a11 = 3 and an (last term) = -1/5 of an A.P.As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore,

a

_{26}= a + (26 – 1)da + 25d = 0 ——–(1)

a11 = a + (11 – 1)d

a + 10d = 3 ———(2)

Solving (1) and (2),

(1) – (2)

a + 25d – (a + 10d) = 0 – 3

15d = -3

d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term is -1/5

5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 – n + 1 = -1

n = 27

Hence, the A.P has 27 terms and its common difference is -1/5.

**Question 12. If the nth term of the A.P. 9, 7, 5, …. is same as the nth term of the A.P. 15, 12, 9, … find n.**

**Solution: **

Given that,

A.P1 = 9, 7, 5, …. and A.P2 = 15, 12, 9, …

nth term of the A.P1 = nth term of the A.P2,

As we know that, to find nth term in an A.P = a + (n – 1)d

For A.P1,

a = 9, d = Second term – first term = 9 – 7 = -2

And, its nth term an = 9 + (n – 1)(-2) = 9 – 2n + 2

an = 11 – 2n ———-(i)

Similarly, for A.P2

a = 15, d = Second term – first term = 12 – 15 = -3

And, its nth term an = 15 + (n – 1)(-3) = 15 – 3n + 3

an = 18 – 3n ——–(ii)

11 – 2n = 18 – 3n

n = 7

Hence, the 7th term of the both the A.Ps are equal.

**Question 13. Find the 12th term from the end of the following arithmetic progressions:**

**(i) 3, 5, 7, 9, …. 201**

**(ii) 3,8,13, … ,253**

**(iii) 1, 4, 7, 10, … ,88**

**Solution: **

In order the find the 12th term from the end of an A.P which has n terms, it is done by simply finding the ((n -12) + 1)th of the A.P.

As we know that, to find nth term in an A.P = a + (n – 1)d

(i)Given that,A.P = 3, 5, 7, 9, …. 201

last term is 201

where, a = 3 and d = (5 – 3) = 2

an = 3 + (n – 1)2 = 201

3 + 2n – 2 = 201

2n = 200

n = 100

Hence the A.P has 100 terms.

therefore, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term.

a

_{89}= 3 + (89 – 1)2= 3 + 88(2)

= 3 + 176 = 179

Hence, the 12th term from the end of the A.P is 179.

(ii)Given that,A.P = 3,8,13, … ,253

last term is 253

where, a = 3 and d = (8 – 3) = 5

an = 3 + (n – 1)5 = 253

3 + 5n – 5 = 253

5n = 253 + 2 = 255

n = 255/5

n = 51

Hence, the A.P has 51 terms.

therefore, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.

a

_{40}= 3 + (40 – 1)5= 3 + 39(5)

= 3 + 195 = 198

Hence, the 12th term from the end of the A.P is 198.

(iii)Given that,A.P = 1, 4, 7, 10, … ,88

where, a = 1 and d = (4 – 1) = 3

last term is 88

an = 1 + (n – 1)3 = 88

1 + 3n – 3 = 8

3n = 90

n = 30

Hence, the A.P has 30 terms.

therefore, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term.

= a

_{89 }= 1 + (19 – 1)3= 1 + 18(3) = 1 + 54 = 55

Hence the 12th term from the end of the A.P is 55.

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