**NCERT Solution For Class 10, Maths, Chapter 5 Arithmetic Progressions, Exercise 5.2** has total 20 questions to discuss. all questions of class 10 maths chapter 5, exercise 5.2 is based on general formula of Arithmetic Progressions.

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Toggle## Class 10, Maths, Chapter 5, Exercise 5.2 Solutions

**Q.1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:**

**Ans:** Blanks may be filled as under:

**(i) a _{n} = a + (n – 1) d**

= 7 + (8-1)3

= (7) +(7×3) = 7+21= 28

**(ii) a _{n} = a + (n – 1) d **

⟹ 0 = -18 + (10 – 1) d

⟹ 18 = 9d

⟹ d = $\frac{18}{9}$= 9

**(iii) a _{n} = a + (n – 1) d **

⟹ -5 = a + (18 – 1) (-3)

⟹ -5 = a + 17 (-3)

⟹ – 5 = a – 51

⟹ a = – 5 + 51 = 46

**(iv) a _{n} = a + (n – 1) d**

⟹ 3.6 = – 18.9 + (n-1) 2.5

⟹ 3.6 + 18.9 = (n – 1) 2.5

⟹ 22.5 = (n – 1) 2.5

⟹ $n-1=\frac{22.5}{2.5}$

⟹ n – 1 = 9

⟹ n = 9 + 1 = 10

**(v) a _{n} = a + (n – 1) d**

a_{n} = 3.5 + (10.5 – 1) × 0

a_{n} = 3.5 + 0 = 3.5

**Q.2. Choose the correct choice in the following and justify:**

**(i) 30 ^{th} term of the AP: 10, 7, 4, ….. is**

**(A) 97 **

**(B) 77 **

**(C) –77 **

**(D) – 87 **

**(ii) 11 ^{th} term of the AP: – 3, –**

**, 2, …., is**

**(A) 28**

**(B) 22**

**(C) –38**

**(D) – 48 $\frac{1}{2}$**

** ****Ans:**

**(i) Correct answer, (C)**

**Justification: **30^{th} term of the AP: 10, 7, 4, ….. can be calculated by:

Here, a = 10, d = 7 – 10 = -3, n = 30^{th} term

We know that, a_{n} = a + (n – 1) d

a_{30} = 10 + (30 – 1) (-3)

a_{30} = 10 + (29)x( -3) = 10 – 87 = -77

So, (C) is the correct choice.

**(ii) Correct answer, (B)**

**Q.3. In the following APs, find the missing terms:**

** (i) 2, ___, 26**

** (ii) ___, 13, ___, 3**

** (iii) 5___, ___, 9$\frac{1}{2}$**

** (iv) -4, ___, ___, ___, ___, 6 **

** (v) ___, 38, ___, ___, ___, -22**

** ****Ans:** **(i) 2, ___, 26 be a, (a + d) and (a+2d)**

∴ a = 2 and (a + 2d) = 26

⟹ (2 + 2d) = 26

⟹ 2d = 26 – 2 = 24

⟹ d = $\frac{24}{2}$=12

Thus, the missing term = a + d = 2 + 12 = 14

**(ii) Let ___, 13, ___, 3 be**

a, (a + d) and (a + 2d) and (a + 3d)

∴ (a + d) = 13 ……(i)

And, (a + 3d) = 3 ……(ii)

Subtracting, (i) from (ii), we get

(a + 3d) – (a + d) = 3 – 13

a + 3d – a – d = -10

2d = -10 ⟹ d = -5

Putting d = -5 in (i), we get

(a + d) = 13

(a – 5) = 13

⟹ a = 13 + 5 = 18

∴ The missing terms are

a =18 and (a+2d) = [18+2(-5)] = 18-10=8

**Q.4. Which term of the AP: 3, 8, 13, 18, . . . , is 78?**

**Ans:** Clearly, the given list is an AP.

We have, a = 3, d = 8- 3 = 5

Let 78 be the n^{th} term of the given AP then,

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

⟹ 78 = 3 + (n-1) 5

⟹ 78 – 3 = 5n – 5

⟹ 75 + 5 = 5n

⟹ 5n = 80

⟹ n = $\frac{80}{5}$ = 16

Thus, 78 is the 16^{th} term of the given list.

**Q.5. Find the number of terms in each of the following APs :**

**(i) 7, 13, 19, . . . , 205 **

** (ii) 18, 15$\frac{1}{2}$, 13, ….., 47**

**Ans: **Clearly, it forms an AP with first term a = 3 and common difference d = 13 – 7 = 6.

**Q.6. Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .**

**Ans:** Here a = 11, d = a_{2} -a_{1 }= 8 – 11= -3

Let – 150 be the n^{th} term of the given AP.

We know that,

a_{n} = a + (n – 1) d

⟹ -150 = 11 + (n – 1) (-3)

⟹ -3 (n-1) = – 150 – 11 = – 161

⟹ n – 1 = $\frac{161}{3}$

⟹ n = $\frac{161}{3}+1=\frac{164}{3}$

Here, the value of n is not an integer. Hence, -150 is not a term of the given AP.

**Q.7. Find the 31 ^{st} term of an AP whose 11^{th} term is 38 and the 16^{th} term is 73.**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference

Now, a_{n} = a + (n – 1) d

∴ a_{11} = a + 10d = 38 ………………… (i)

And, a_{16} = a + 15d = 73 ………………… (ii)

Subtracting (i) from (ii), we get

5d = 35 ⟹ d == 7

And then from (i),

a + 10d = 38

a + 10 × 7 =38

⟹ a = 38 – 70 = -32

Now, we have a = -32 and d = 7

∴ a_{31} = a + (31 – 1) d

a_{31} = (-32) + (30×7) = -32 + 210 = 178

Hence, **the 31 ^{st} term is 178.**

**Q.8. An AP consists of 50 terms of which 3 ^{rd} term is 12 and the last term is 106. Find the 29^{th} term.**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference

Now, a_{n} = a + (n – 1) d

∴ a_{3} = a + 2d = 12 …………… (i)

And, a_{50} = a + 49d = 106 …………… (ii)

Subtracting (i) and (ii), we get

47 d = 94 ⟹ d = $\frac{94}{47}$=2

Putting value of d in (i),

a + 2d = 12

a + (2 × 2) = 12

⟹ a = 12 – 4 = 8

Now, we have a = 8 and d = 2

∴ a_{29} = a + (29 – 1) d

a_{29} = 8 + (28 x 2)

a_{29} = 8 + 56 = 64

Hence, **the 29 ^{st} term is 64.**

**Q.9. If the 3 ^{rd} and the 9^{th} terms of an AP are 4 and – 8 respectively, which term of this AP is zero?**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference

∴ a_{3} = a + 2d = 4 ………… (i)

And, a_{9} = a + 8d = – 8 ………… (ii)

Subtracting (i) and (ii), we get

6d= -12 ⟹ d = $\frac{-12}{6}$ = -2

Putting the value of d in eq.(i),

a + 2d = 4

a + 2× (-2) = 4

⟹ a = 4 + 4 = 8

Now, we have, a = 8 and d = -2

Let a_{n} = 0

⟹ a_{n} = a + (n – 1) d

⟹ 0 = 8 + (n – 1) (-2)

⟹ (n -1) (-2) = – 8

⟹ n – 1 =$\frac{-8}{-2}$ = 4

⟹ n = 4 + 1 = 5

Thus, 5^{th} term of the given AP is zero.

**Q.10. The 17 ^{th} term of an AP exceeds its 10^{th} term by 7. Find the common difference.**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference.

It is given a_{17} – a_{10} = 7

⟹ (a + 16d) – (a + 9d) = 7

⟹ 7d = 7

⟹ d = 1

Hence, the common difference is 1.

**Q.11. Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54 ^{th} term?**

**Ans:** Here, a = 3, d = a_{2} – a_{1 }= 15- 3 = 12. Then,

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

a_{54} = 3 + (54 -1) x12

a_{54} = 3 + 53 x 12

a_{54} = 3 + 636 = 639

Let a_{n} be 132 more than its 54^{th} term

i.e., a_{n} = a_{54} + 132 = 639 + 132 = 771

⟹ 771 = a + (n – 1) d

⟹ 771 = 3 + (n – 1) 12

⟹ (n–1)12 =771-3

⟹ (n–1) = $\frac{768}{12}$ = 64

⟹ n = 64 + 1 = 65

Then, 65^{th} term is 132 more than its 54^{th} term.

**Q.12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?**

**Ans:** Let ‘a’ and ‘b’ the first term of two APs and ‘d’ is the common difference.

Given, a_{100} – b_{100} = 100

⟹ [a + (n – 1) d] – [b + (n – 1) d] = 100

⟹ (a – b) = 100 ……… (i)

Now, let the difference of their 1000^{th} terms are ‘x’. then,

a_{1000} – b_{1000} = x ……… (ii)

⟹ [a + (n – 1) d] – [b + (n – 1) d] = x

⟹ (a – b) = x ……… (iii)

From (i) and (iii),

x = 100

Putting the value of ‘x’ in eq.(ii)

⟹ a_{1000} – b_{1000} = x

⟹ a_{1000} – b_{1000} = 100

Hence, the difference between 1000^{th} terms is the same as the difference between 100^{th} terms, i.e., 100.

**Q.13. How many three-digit numbers are divisible by 7?**

**Ans:** 994 is the last 3-digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …, 994.

Here, a = 105, d = a_{2} – a_{1}= 112 – 105 = 7, a_{n}= 994 (last term) and n =? (no. of terms)

We know that, a_{n} = a + (n- 1) d

⟹ 994 = 105 + (n – 1)7

⟹ 7(n – 1) = 994 – 105

⟹ 7 (n – 1) = 889

⟹ n – 1 = $\frac{889}{7}$ = 127

⟹ n = 127 + 1 = 128

Hence, there are 128 numbers of three digit which are divisible by 7.

**Q.14. How many multiples of 4 lie between 10 and 250?**

**Ans:** The multiple of 4 between 10 and 250 are 12,16,20,24…. 248.

Here, a = 12, d = a_{2} – a_{1}= 16 – 12 = 4 and a_{n} = 248

Let there be n terms in the AP Then,

∴ a_{n} = a + (n-1) d

⟹ 248 = 12 + (n -1) 4

⟹ 4 (n – 1) = 248 – 12

⟹ 4 (n – 1) = 236

⟹ n – 1 = $\frac{236}{4}$ = 59

⟹ n = 59 + 1 = 60

Hence, there are 60 multiples of 4 between 10 and 250.

**Explanation:**

We observe that 12 is the first integer between 10 and 250, which is a multiple of 4 (i.e., divisible by 4). Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 - 2 = 248 is the largest integer divisible by 4(i.e., multiple of 4) and lying between 10 and 250. Thus, we have to find the number of terms in an AP with first term = 12, last term = 248, and common difference = 4 (as the number are divisible by 4).

**Q.15. For what value of n, are the n ^{th} terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . equal?**

**Ans:** It n^{th} terms of the APs 63, 65, 67, … and 3, 10, 17, … are equal. Then

For 1^{st} AP, a = 63, d = a_{2} – a_{1}= 65 – 63 = 2

And, For 2^{nd} AP, a = 3, d = a_{2} – a_{1}= 10 – 3 = 7

According to question, n^{th} term of both are equal. Therefore,

a + (n-1) d (For 1^{st}) = a + (n-1) d (For 2^{nd})

⟹ 63 + (n – 1) 2 = 3 + (n – 1) 7

⟹ (n -1)7 – (n –1)2 = 63 – 3

⟹ (n -1) (7 –2) = 63 – 3

⟹ 5 (n-1) = 60

** **⟹ n – 1 = $\frac{60}{5}$ = 12

⟹** **n = 12 + 1 = 13

Hence, the 13^{th} terms of the two given APs are equal.

**Q.16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference

Here, a_{3} = 16 and a_{7} – a_{5} =12

⟹ a_{3} =16 = a + (3-1) d

⟹ a + 2d = 16 ……………….. (i)

And, a + (7-1) d – [a + (5-1) d] = 12

(a + 6d) – (a + 4d) = 12

⟹ 2d = 12

⟹ d = 6 …………….. (ii)

Putting the value of d in eq.(i), we get

a + 2 × 6 = 16

⟹ a = 16 – 12 = 4

Now, we have, a = 4, d = 6.

∴ a_{n} = a + (n-1) d

a_{1} = 4 + (1-1) 6 = 4

a_{2} = 4 + (2-1) 6 = 10

a_{3} = 4 + (3-1) 6 = 16

a_{4} = 4 + (4-1) 6 = 22

So, the required AP is 4, 10, 16, 22, …….

**Q.17. Find the 20 ^{th} term from the last term of the AP : 3, 8, 13, . . ., 253.**

**Ans:** Given AP: **3, 8, 13, . . ., 253**

Here, d = a_{2} – a_{1 }= 8 – 3 = 5

To find the 20^{th} term from the last, we are reversing the given AP by using the common difference, therefore,

AP: 253, 248, 243, ……….13,8,3.

Now, we have a = 253, d = a_{2} – a_{1 }= 248 – 253 = -5 and a_{n} = 20^{th} term

We know that,

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

⟹ a_{20} = 253 + (20-1) (-5)

⟹ a_{20} = 253 + (19) (-5)

⟹ a_{20} = 253 – 95

⟹ a_{20} = 158

**Hence, the 20 ^{th} term from the last term is 158.**

we have, l = Last term = 253

And, d = Common difference = 8 – 3 = 5

∴ 20^{th} term from the end = l – (20-1) d

= l – 19d

= 253 – 19 × 5

= 253 – 95 = 158

**Q.18. The sum of the 4 ^{th} and 8^{th} terms of an AP is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the AP.**

**Ans: **Let ‘a’ be the first term and ‘d’ the common difference

Here, a_{4} + a_{8} = 24

⟹ [a+(4-1) d] + [a+(8-1) d] = 24

⟹ (a + 3d) + (a + 7d) = 24

⟹ 2a + 10d = 24

⟹ 2(a + 5d) = 24

⟹ a + 5d = 12 …………………(i)

and, a_{6} + a_{10} = 44

⟹ [a+(6-1) d] + [a+(10-1) d] = 44

⟹ (a + 5d) + (a + 9d) = 44

⟹ 2a + 14d = 44

⟹ 2(a + 7d) = 44

⟹ a + 7d = 22 …………………(ii)

Subtracting (i) from (ii), we get

2d = 10

⟹ d = 5

Putting the value of d in (i),

a + 25 = 12

⟹ a = – 13

Now, we have a = -13 and d = 5. Then, the first three terms are

∴ a_{n} = a + (n-1) d

a_{1} = (-13) + (1-1) 5 = -13

a_{2} = (-13) + (2-1) 5 = -8

a_{3} = (-13) + (3-1) 5 = -3

hence, the first three terms are, -13,-8,-3.

**Q.19. Subba Rao started work in 1995 at an annual salary of ****₹****5000 and received an increment of ****₹****200 each year. In which year did his income reach ****₹**** 7000?**

**Ans:** The annual salary drawn by Subba Rao in the years 1995, 1996, 1997, etc. is Rs 5,000, Rs 5,200, Rs 5,400, …, Rs 7,000

The list of these numbers are 5000, 5200, 5400, …….., 7000.

It forms an AP. [∵ a_{2} – a_{1} = a_{3} – a_{2} = 200]

Let after ‘n’ years his salary will be Rs 7000

∴ a_{n} = 7000 = a + (n-1) d

⟹** **7000 = a + (n-1) d

⟹ 7000 = 5000 + (n-1) (200)

⟹ 200 (n-1) = 7000 – 5000

⟹ n – 1 = $\frac{2000}{200}$ = 10

⟹ n = 10 + 1= 11

Now, to find required year, we have AP of years: 1995, 1996, 1997 …

Here, a = 1995, d = a_{2} – a_{1}= 1996 – 1995 = 1

∴ a_{n} = a + (n-1) d

a_{11} = 1995 + (11-1) x1 = 1995 + 10 = 2005

Thus, in the 11^{th} year (i.e., in 2005) of his service, Subba Rao drew an annual salary of Rs 7,000.

**Q.20. Ramkali saved ****₹****5 in the first week of a year and then increased her weekly savings by ****₹****1.75. If in the n ^{th} week, her weekly savings become **

**₹**

**20.75, find ‘n’.**

**Ans:** Ramkali’s savings in the subsequent weeks are respectively Rs 5, Rs (5 + 1.75), Rs(5 +2×1.75), Rs (5 + 3×1.75), …

So it is a form of AP: 5, 6.75,8.5, 10.25,…..

Here, we have, a = 5, d = a_{2} – a_{1} = 6.75 – 5 = 1.75

And, in the n^{th} week her saving will be Rs 20.75

∴ a_{n} = 20.75 = a + (n-1) d

⟹ 20.75 = 5 + (n-1) × 1.75

⟹ (n-1) × 1.75 = 20.75 – 5 = 15.75

⟹ n – 1 = $\frac{15.75}{1.75}$ = 9

⟹ n = 9 +1 = 10

Hence, in 10^{th} week, Ramkali savings become ₹20.75.