The concepts of the general term, nth term from the end and middle term of an A.P are testified in this exercise. As these are the main parts of the chapter, we strongly recommend students to refer the RD Sharma Solutions Class 10 prepared by subject experts at BYJUâ€™S as its a very useful resource for students to prepare well for their exams. Students can use RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.4 PDF for getting a clear understanding of the problems.

## RD Sharma Solutions for Class 10 Chapter 9 Arithmetic Progressions Exercise 9.4 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 9 Arithmetic Progressions Exercise 9.4

**1. Find:**

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

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

**(iii) n ^{th}Â term of the AP 13, 8, 3, -2, ……….Â **

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

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

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

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

**Solution:**

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

First term (a) = 1

Common difference (d) = Second term – First term

= 4 – 1 = 3.

We know that, n^{th}Â term in an A.P = a + (n – 1)d

Then, 10^{th}Â term in the A.P is 1 + (10 – 1)3

= 1 + 9×3

= 1 + 27

= 28

âˆ´ 10^{th}Â term of A. P. is 28

(ii) Given A.P. is âˆš2, 3âˆš2, 5âˆš2, â€¦â€¦.

First term (a) = âˆš2Â

Common difference = Second term â€“ First term

= 3âˆš2 – âˆš2

â‡’ d = 2âˆš2Â

We know that, n^{th}Â term in an A. P. = a + (n – 1)d

Then, 18^{th}Â term of A. P. = âˆš2 + (18 – 1)2âˆš2

= âˆš2 + 17.2âˆš2

= âˆš2 (1+34)

= 35âˆš2

âˆ´ 18^{th}Â term of A. P. is 35âˆš2

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

First term (a) = 13

Common difference (d) = Second term first term

= 8 – 13 = â€“ 5

We know that, n^{th}Â term of an A.P. a_{n}Â = a +(n – 1)d

= 13 + (n – 1) – 5

= 13 – 5n + 5

âˆ´ n^{th} term of the A.P is a_{n}Â = 18 – 5n

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

First term (a) = -40

Common difference (d) = Second term – fast term

= -15 – (- 40)

= 40 – 15

= 25

We know that, n^{th}Â term of an A.P. a_{n}Â = a + (n – 1)d

Then, 10^{th}Â term of A. P. a_{10}Â = -40 + (10 – 1)25

= â€“ 40 + 9.25

= â€“ 40 + 225

= 185

âˆ´ 10^{th}Â term of the A. P. is 185

(v) Given sequence is 117, 104, 91, 78, ………….Â

First term (a) = 117

Common difference (d) = Second term – first term

= 104 – 117

= â€“ 13

We know that, n^{th}Â term = a + (n – 1)d

Then, 8^{th}Â term = a + (8 – 1)d

= 117 + 7(-13)

= 117 – 91

= 26

âˆ´ 8^{th}Â term of the A. P. is 26

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

First term (a) = 10.0

Common difference (d) = Second term – first term

= 10.5 – 10.0 = 0.5

We know that, n^{th}Â term a_{n}Â = a + (n – 1)d

Then, 11^{th}Â term a_{11}Â = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5

= 10.0 + 5

=15.0

âˆ´ 11^{th}Â term of the A. P. is 15.0

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

First term (a) = 3/4

Common difference (d) = Second term – first term

= 5/4 – 3/4

= 2/4

We know that, n^{th}Â term a_{n}Â = a + (n – 1)d

Then, 9^{th}Â term a_{9}Â = a + (9 – 1)d

âˆ´ 9^{th}Â term of the A. P. is 19/4.

**Â **

**2.(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 A.P. is 3, 8, 13, ………..

First term (a) = 3

Common difference (d) = Second term – first term

= 8 – 3

= 5

We know that, n^{th}Â term (a_{n}) = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 248

248 = 3+(n – 1)5

248 = -2 + 5n

5n = 250

n =250/5 = 50

âˆ´ 50^{th}Â term in the A.P is 248.

(ii) Given A. P is 84, 80, 76, …………Â

First term (a) = 84

Common difference (d) = a_{2}Â – a

= 80 – 84

= â€“ 4

We know that, n^{th}Â term (a_{n}) = a +(n – 1)d

And, given nth term is 0

0 = 84 + (n – 1) – 4

84 = +4(n – Â 1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

âˆ´ 22^{nd}Â term in the A.P is 0.

(iii) Given A. P 4, 9, 14, …………Â

First term (a) = 4

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

= 9 – 4

= 5

We know that, n^{th}Â term (a_{n}) = a + (n – 1)d

And, given n^{th}Â term is 254

4 + (n – 1)5 = 254

(n – 1)âˆ™5 = 250

n – 1 = 250/5 = 50Â

n = 51

âˆ´Â 51^{th}Â term in the A.P is 254.Â

(iv) Given A. P 21, 42, 63, 84, ………Â

a = 21, d = a_{2}Â â€“ a_{1}

= 42 – 21

= 21

We know that, n^{th}Â term (a_{n}) = a +(n – 1)d

And, given n^{th} term = 420

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

âˆ´ 20^{th}Â term is 420.

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

Fiat term (a) = 121

Common difference (d) = 117 – 121

= – 4

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, for some n^{th}Â term is negative i.e., a_{n}Â < 0

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.

âˆ´ 32^{nd}Â term in the A.P will be the first negative term.

**3.(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€¦. ?**

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

**Solutions: **

(i) Given, A.P. 7, 10, 13,â€¦

Here, a = 7 and d = a_{2} â€“ a_{1} = 10 â€“ 7 = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = 68

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n â€“ 3 = 68

3n + 4 = 68

3n = 64

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

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

(ii) Given, A.P. 3, 8, 13,â€¦

Here, a = 3 and d = a_{2} â€“ a_{1} = 8 â€“ 3 = 5

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = 302

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n â€“ 5 = 302

5n – 2 = 302

5n = 304

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

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

(iii) Given, A.P. 11, 8, 5, 2, â€¦

Here, a = 11 and d = a_{2} â€“ a_{1} = 8 â€“ 11 = -3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = -150

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 â€“ 3n + 3 = -150

3n = 150 + 14

3n = 164

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

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

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

**(i) 7, 10, 13, â€¦.., 43**

**(ii) -1, -5/6, -2/3, -1/2, â€¦ , 10/3**

**(iii) 7, 13, 19, â€¦, 205 **

**(iv) 18, 15Â½, 13, â€¦., -47**

**Solution: **

(i) Given, A.P. 7, 10, 13, â€¦.., 43

Here, a = 7 and d = a_{2} â€“ a_{1} = 10 â€“ 7 = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 43

a + (n – 1)d = 43

7 + (n – 1)(3) = 43

7 + 3n â€“ 3 = 43

3n = 43 â€“ 4

3n = 39

â‡’ n = 13

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

(ii) Given, A.P. -1, -5/6, -2/3, -1/2, â€¦ , 10/3

Here, a = -1 and d = a_{2} â€“ a_{1} = -5/6 â€“ (-1) = 1/6

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 10/3

a + (n – 1)d = 10/3

-1 + (n – 1)(1/6) = 10/3

-1 + n/6 â€“ 1/6 = 10/3

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

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

n = (20 + 6 + 1)

â‡’ n = 27

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

(iii) Given, A.P. 7, 13, 19, â€¦, 205

Here, a = 7 and d = a_{2} â€“ a_{1} = 13 â€“ 7 = 6

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 205

a + (n – 1)d = 205

7 + (n – 1)(6) = 205

7 + 6n â€“ 6 = 205

6n = 205 â€“ 1

n = 204/6

â‡’ n = 34

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

(iv) Given, A.P. 18, 15Â½, 13, â€¦., -47

Here, a = 7 and d = a_{2} â€“ a_{1} = 15Â½ â€“ 18 = 5/2

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = -47

a + (n – 1)d = 43

18 + (n – 1)(-5/2) = -47

18 â€“ 5n/2 + 5/2 = -47

36 â€“ 5n + 5 = -94

5n = 94 + 36 + 5

5n = 135

â‡’ n = 27

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

**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,

a = 5 and d = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So, for the given A.P. a_{n} = 5 + (n – 1)3 = 3n + 2

Also given, last term = 80

â‡’ 3n + 2 = 80

3n = 78

n = 78/3 = 26

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

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

**Solution:**

Given,

a_{6} = 19 and a_{17} = 41

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So,

a_{6} = a + (6-1)d

â‡’ a + 5d = 19 â€¦â€¦ (i)

Similarity,

a_{17} = a + (17 – 1)d

â‡’ a + 16d = 41 â€¦â€¦ (ii)

Solving (i) and (ii),

(ii) â€“ (i) â‡’

a + 16d â€“ (a + 5d) = 41 â€“ 19

11d = 22

â‡’ d = 2

Using d in (i), we get

a + 5(2) = 19

a = 19 â€“ 10 = 9

Now, the 40^{th} term is given by a_{40} = 9 + (40 – 1)2 = 9 + 78 = 87

Therefore the 40^{th} term is 87.

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

** Solution: **

Given,

a_{9} = 0

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So, a + (9 – 1)d = 0 â‡’ a + 8d = 0 â€¦â€¦(i)

Now,

29^{th} term is given by a_{29} = a + (29 – 1)d

â‡’ a_{29} = a + 28d

And, a_{29} = (a + 8d) + 20d [using (i)]

â‡’ a_{29} = 20d â€¦.. (ii)

Similarly, 19^{th} term is given by a_{19} = a + (19 – 1)d

â‡’ a_{19} = a + 18d

And, a_{19} = (a + 8d) + 10d [using (i)]

â‡’ a_{19} = 10d â€¦..(iii)

On comparing (ii) and (iii), itâ€™s clearly seen that

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

Therefore, 29^{th} term is double the 19^{th} term.

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

**Solution: **

Given,

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

We know that, n^{th}Â term a_{n} = a + (n – 1)d

â‡’ 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 â€“ 1)d = 0

â‡’ a_{25} = 0

Therefore, the 25^{th} term of the A.P. is zero.

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

**Solution: **

Given,

A_{10} = 41 and a_{18} = 73

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So,

a_{10} = a + (10 – 1)d

â‡’ a + 9d = 41 â€¦â€¦ (i)

Similarity,

a_{18} = a + (18 – 1)d

â‡’ a + 17d = 73 â€¦â€¦ (ii)

Solving (i) and (ii),

(ii) â€“ (i) â‡’

a + 17d â€“ (a + 9d) = 73 â€“ 41

8d = 32

â‡’ d = 4

Using d in (i), we get

a + 9(4) = 41

a = 41 â€“ 36 = 5

Now, the 26^{th} term is given by a_{26} = 5 + (26 – 1)4 = 5 + 100 = 105

Therefore the 26^{th} term is 105.

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

**Solution: **

Given,

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

We know that, n^{th}Â term a_{n} = a + (n – 1)d

â‡’ a_{24} = 2(a_{10})

a + (24 – 1)d = 2(a + (10 – 1)d)

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

a + 23d = 2a + 18d

a = 5d â€¦. (1)

Now, the 72^{nd} 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(a_{34})

â‡’ a_{72 }= 2(a_{34})

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

**11. The 26 ^{th}, 11^{th} 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,

a_{26 }= 0 , a_{11 }= 3 and a_{n }(last term) = -1/5 of an A.P.

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Then,

a_{26}^{ }= a + (26 – 1)d

â‡’ a + 25d = 0 â€¦..(1)

And,

a_{11}^{ }= a + (11 – 1)d

â‡’ a + 10d = 3 â€¦â€¦ (2)

Solving (1) and (2),

(1) â€“ (2) â‡’

a + 25d â€“ (a + 10d) = 0 â€“ 3

15d = -3

â‡’ d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term a_{n} = -1/5

â‡’ 5 + (n – 1)(-1/5) = -1/5

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

25 â€“ n + 1 = -1

n = 27

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

**12. If the n ^{th} term of the A.P. 9, 7, 5, â€¦. is same as the n^{th} term of the A.P. 15, 12, 9, â€¦ find n. **

**Solution: **

Given,

A.P_{1 }= 9, 7, 5, â€¦. and A.P_{2 }= 15, 12, 9, â€¦

And, we know that, n^{th}Â term a_{n} = a + (n – 1)d

For A.P_{1},

a = 9, d = Second term â€“ first term = 9 â€“ 7 = -2

And, its n^{th}Â term a_{n }= 9 + (n – 1)(-2) = 9 â€“ 2n + 2

a_{n }= 11 â€“ 2n â€¦..(i)

Similarly, for A.P_{2}

a = 15, d = Second term â€“ first term = 12 â€“ 15 = -3

And, its n^{th}Â term a_{n }= 15 + (n – 1)(-3) = 15 â€“ 3n + 3

a_{n }= 18 – 3n â€¦..(ii)

According to the question, its given that

n^{th} term of the A.P_{1} = n^{th} term of the A.P_{2}

â‡’ 11 â€“ 2n = 18 – 3n

n = 7

Therefore, the 7^{th} term of the both the A.Ps are equal.

**13. Find the 12 ^{th} term from the end of the following arithmetic progressions: **

**(i) 3, 5, 7, 9, â€¦. 201**

**(ii) 3,8,13, â€¦ ,253**

**(iii) 1, 4, 7, 10, â€¦ ,88**

**Solution: **

In order the find the 12^{th} term for the end of an A.P. which has n terms, its done by simply finding the ((n -12) + 1)^{th} of the A.P

And we know, n^{th}Â term a_{n} = a + (n – 1)d

(i) Given A.P = 3, 5, 7, 9, â€¦. 201

Here, a = 3 and d = (5 – 3) = 2

Now, find the number of terms when the last term is known i.e, 201

a_{n} = 3 + (n – 1)2 = 201

3 + 2n â€“ 2 = 201

2n = 200

n = 100

Hence, the A.P has 100 terms.

So, the 12^{th} term from the end is same as (100 â€“ 12 + 1)^{th} of the A.P which is the 89^{th} term.

â‡’ a_{89 }= 3 + (89 – 1)2

= 3 + 88(2)

= 3 + 176

= 179

Therefore, the 12^{th} term from the end of the A.P is 179.

(ii) Given A.P = 3,8,13, â€¦ ,253

Here, a = 3 and d = (8 – 3) = 5

Now, find the number of terms when the last term is known i.e, 253

a_{n} = 3 + (n – 1)5 = 253

3 + 5n â€“ 5 = 253

5n = 253 + 2 = 255

n = 255/5

n = 51

Hence, the A.P has 51 terms.

So, the 12^{th} term from the end is same as (51 â€“ 12 + 1)^{th} of the A.P which is the 40^{th} term.

â‡’ a_{40 }= 3 + (40 – 1)5

= 3 + 39(5)

= 3 + 195

= 198

Therefore, the 12^{th} term from the end of the A.P is 198.

(iii) Given A.P = 1, 4, 7, 10, â€¦ ,88** **

Here, a = 1 and d = (4 – 1) = 3

Now, find the number of terms when the last term is known i.e, 88

a_{n} = 1 + (n – 1)3 = 88

1 + 3n â€“ 3 = 88

3n = 90

n = 30

Hence, the A.P has 30 terms.

So, the 12^{th} term from the end is same as (30 â€“ 12 + 1)^{th} of the A.P which is the 19^{th} term.

â‡’ a_{89 }= 1 + (19 – 1)3

= 1 + 18(3)

= 1 + 54

= 55

Therefore, the 12^{th} term from the end of the A.P is 55.

**14. The 4 ^{th} term of an A.P. is three times the first and the 7^{th} term exceeds twice the third term by 1. Find the first term and the common difference.**

**Solution: **

Letâ€™s consider the first term and the common difference of the A.P to be a and d respectively.

Then, we know that a_{n} = a + (n – 1)d

Given conditions,

4^{th} term of an A.P. is three times the first

Expressing this by equation we have,

â‡’ a_{4} = 3(a)

a + (4 – 1)d = 3a

3d = 2a â‡’ a = 3d/2â€¦â€¦.(i)

And,

7^{th} term exceeds twice the third term by 1

â‡’ a_{7} = 2(a_{3}) + 1

a + (7 â€“ 1)d = 2(a + (3â€“1)d) + 1

a + 6d = 2a + 4d + 1

a â€“ 2d +1 = 0 â€¦.. (ii)

Using (i) in (ii), we have

3d/2 â€“ 2d + 1 = 0

3d â€“ 4d + 2 = 0

d = 2

So, putting d = 2 in (i), we get a

â‡’ a = 3

Therefore, the first term is 3 and the common difference is 2.

**15. Find the second term and the n ^{th} term of an A.P. whose 6^{th} term is 12 and the 8^{th} term is 22. **

**Solution: **

Given, in an A.P

a_{6} = 12 and a_{8} = 22

We know that a_{n} = a + (n – 1)d

So,

a_{6 }= a + (6-1)d = a + 5d = 12 â€¦. (i)

And,

a_{8} = a + (8-1)d = a + 7d = 22 â€¦â€¦. (ii)

Solving (i) and (ii), we have

(ii) – (i) â‡’

a + 7d â€“ (a + 5d) = 22 â€“ 12

2d = 10

d = 5

Putting d in (i) we get,

a + 5(5) = 12

a = 12 â€“ 25

a = -13

Thus, for the A.P: a = -13 and d = 5

So, the n^{th} term is given by a_{n} = a + (n-1)d

a_{n }= -13 + (n-1)5 = -13 + 5n â€“ 5

â‡’ a_{n }= 5n â€“ 18

Hence, the second term is given by a_{2 }= 5(2) â€“ 18 = 10 â€“ 18 = -8

**16. How many numbers of two digit are divisible by 3? **

**Solution: **

The first 2 digit number divisible by 3 is 12. And, the last 2 digit number divisible by 3 is 99.

So, this forms an A.P.

12, 15, 18, 21, â€¦. , 99

Where, a = 12 and d = 3

Finding the number of terms in this A.P

â‡’ 99 = 12 + (n-1)3

99 = 12 + 3n â€“ 3

90 = 3n

n = 90/3 = 30

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

**17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32 ^{nd} term.**

**Solution: **

Given, an A.P of 60 terms

And, a = 7 and a_{60} = 125

We know that a_{n} = a + (n – 1)d

â‡’ a_{60} = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

So, the 32^{nd} term is given by

a_{32} = 7 + (32 -1)2 = 7 + 62 = 69

â‡’ a_{32} = 69

**18. The sum of 4 ^{th} and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34. Find the first term and the common difference of the A.P.**

**Solution: **

Given, in an A.P

The sum of 4^{th} and 8^{th} terms of an A.P. is 24

â‡’ a_{4} + a_{8} = 24

And, we know that a_{n} = a + (n – 1)d

2a + 10d = 24

a + 5d = 12 â€¦. (i)

Also given that,

the sum of the 6^{th} and 10^{th} terms is 34

â‡’ a_{6} + a_{10} = 34

2a + 14d = 34

a + 7d = 17 â€¦â€¦ (ii)

Subtracting (i) form (ii), we have

a + 7d â€“ (a + 5d) = 17 â€“ 12

2d = 5

d = 5/2

Using d in (i) we get,

a + 5(5/2) = 12

a = 12 â€“ 25/2

a = -1/2

Therefore, the first term is -1/2 and the common difference is 5/2.

**19. The first term of an A.P. is 5 and its 100 ^{th} term is -292. Find the 50^{th} term of this A.P. **

**Solution: **

Given, an A.P whose

a = 5 and a_{100} = -292

We know that a_{n} = a + (n – 1)d

a100 = 5 + 99d = -292

99d = -297

d = -3

Hence, the 50^{th} term is

a_{50} = a + 49d = 5 + 49(-3) = 5 â€“ 147 = -142

**20. Find a _{30} â€“ a_{20} for the A.P.**

**(i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, â€¦â€¦**

**Solution: **

We know that a_{n} = a + (n – 1)d

So, a_{30} â€“ a_{20} = (a + 29d) â€“ (a + 19) =10d

(i) Given A.P. -9, -14, -19, -24

Here, a = -9 and d = -14 â€“ (-9) = = -14 + 9 = -5

So, a_{30} â€“ a_{20} = 10d

= 10(-5)

= -50

(ii) Given A.P. a, a+d, a+2d, a+3d, â€¦â€¦

So, a_{30} â€“ a_{20} = (a + 29d) â€“ (a + 19d

=10d

**21. Write the expression a _{n} â€“ a_{k} for the A.P. a, a+d, a+2d, â€¦..**

**Hence, find the common difference of the A.P. for which**

**(i) 11 ^{th} term is 5 and 13^{th} term is 79.**

**(ii) a _{10 }â€“ a_{5} = 200 **

**(iii) 20 ^{th} term is 10 more than the 18^{th} term. **

**Solution: **

Given A.P. a, a+d, a+2d, â€¦..

So, a_{n }= a + (n-1)d = a + nd â€“d

And, a_{k }= a + (k-1)d = a + kd â€“ d

a_{n }– a_{k }= (a + nd â€“ d) â€“ (a + kd â€“ d)

= (n â€“ k)d

(i) Given 11^{th} term is 5 and 13^{th} term is 79,

Here n = 13 and k = 11,

a_{13 }â€“ a_{11 }= (13 â€“ 11)d = 2d

â‡’ 79 â€“ 5 = 2d

d = 74/2 = 37

(ii) Given, a_{10 }â€“ a_{5} = 200

â‡’ (10 – 5)d = 200

5d = 200

d = 40

(iii) Given, 20^{th} term is 10 more than the 18^{th} term.

â‡’ a_{20} â€“ a_{18} = 10

(20 – 18)d = 10

2d = 10

d = 5

**22. Find n if the given value of x is the n ^{th} term of the given A.P. **

**(i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, â€¦; x = -151**

**(iii) 5Â½, 11, 16Â½, 22, â€¦.; x = 550 (iv) 1, 21/11, 31/11, 41/11, â€¦; x = 171/11**

**Solution: **

(i) Given A.P. 25, 50, 75, 100, â€¦â€¦ ,1000

Here, a = 25 d = 50 â€“ 25 = 25

Last term (n^{th} term) = 1000

We know that a_{n} = a + (n – 1)d

â‡’ 1000 = 25 + (n-1)25

1000 = 25 + 25n â€“ 25

n = 1000/25

n = 40

(ii) Given A.P. -1, -3, -5, -7, â€¦., -151

Here, a = -1 d = -3 â€“ (-1) = -2

Last term (n^{th} term) = -151

We know that a_{n} = a + (n – 1)d

â‡’ -151 = -1 + (n-1)(-2)

-151 = -1 – 2n + 2

n = 152/2

n = 76

(iii) Given A.P. 5Â½, 11, 16Â½, 22, â€¦ , 550

Here, a = 5Â½ d = 11 â€“ (5Â½) = 5Â½ = 11/2

Last term (n^{th} term) = 550

We know that a_{n} = a + (n – 1)d

â‡’ 550 = 5Â½ + (n-1)(11/2)

550 x 2 = 11+ 11n â€“ 11

1100 = 11n

n = 100

(iv) Given A.P. 1, 21/11, 31/11, 41/11, 171/11

Here, a = 1 d = 21/11 â€“ 1 = 10/11

Last term (n^{th} term) = 171/11

We know that a_{n} = a + (n – 1)d

â‡’ 171/11 = 1 + (n-1)10/11

171 = 11 + 10n â€“ 10

n = 170/10

n = 17

**23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15 ^{th} term. **

**Solution: **

Given, an A.P in which,

a_{8} = 1/2(a_{2})

a_{11} = 1/3(a_{4}) + 1

We know that a_{n} = a + (n – 1)d

â‡’ a_{8} = 1/2(a_{2})

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 â€¦â€¦ (i)

And, a_{11} = 1/3(a_{4}) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 â€¦â€¦ (ii)

Solving (i) and (ii), by (ii) â€“ 2x(i) â‡’

2a + 27d â€“ 2(a + 13d) = 3 – 0

d = 3

Putting d in (i) we get,

a + 13(3) = 0

a = -39

Thus, the 15^{th} term a_{15 }= -39 + 14(3) = -39 + 42 = 3

**24. Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12. **

**Solution: **

Given, in an A.P

a_{3} = 16 and a_{7} = a_{5} + 12

We know that a_{n} = a + (n – 1)d

â‡’ a + 2d = 16â€¦â€¦ (i)

And,

a + 6d = a + 4d + 12

2d = 12

â‡’ d = 6

Using d in (i), we have

a + 2(6) = 16

a = 16 â€“ 12 = 4

Hence, the A.P is 4, 10, 16, 22, â€¦â€¦.

**25. The 7 ^{th} term of an A.P. is 32 and its 13^{th} term is 62. Find the A.P.**

**Solution: **

Given,

a_{7 }= 32 and a_{13 }= 62

From a_{n }– a_{k }= (a + nd â€“ d) â€“ (a + kd â€“ d)

= (n â€“ k)d

a_{13} â€“ a_{7} = (13 – 7)d = 62 â€“ 32 = 30

6d = 30

d = 5

Now,

a_{7} = a + (7 – 1)5 = 32

a + 30 = 32

a = 2

Hence, the A.P is 2, 7, 12, 17, â€¦â€¦

**26. Which term of the A.P. 3, 10, 17, â€¦. will be 84 more than its 13 ^{th} term ?**

**Solution: **

Given, A.P. 3, 10, 17, â€¦.

Here, a = 3 and d = 10 â€“ 3 = 7

According the question,

a_{n} = a_{13} + 84

Using a_{n} = a + (n – 1)d,

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n â€“ 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

n = 25

Therefore, it the 25^{th} term which is 84 more than its 13^{th} term.

**27. Two arithmetic progressions have the same common difference. The difference between their 100 ^{th} terms is 100, what is the difference between their 1000^{th} terms?**

**Solution: **

Let the two A.Ps be A.P_{1} and A.P_{2}

For A.P_{1} the first term = a and the common difference = d

And for A.P_{2} the first term = b and the common difference = d

So, from the question we have

a_{100} â€“ b_{100} = 100

(a + 99d) â€“ (b + 99d) = 100

a – b = 100

Now, the difference between their 1000^{th} terms is,

(a + 999d) â€“ (b + 999d) = a â€“ b = 100

Therefore, the difference between their 1000^{th} terms is also 100.