# NCERT Class 10 Math Chapter 5 Important Questions Answer – Arithmetic Progressions

 Class 10 Chapter Arithmetic Progressions Subject Math Category Important Question Answer

## Class 10 Math Chapter 5 Important Question Answer

Q1. If the sum of first 7 terms of A. P. is 49 and sum of first 17 terms is 289, then find the sum of n terms of A. P. Most Important

Ans. Given S7 = 49 and S17 = 289, Sn = ?

Using Sn = [2a + (n – 1)d]

S7 = [2a + 6d] =    49  → 2a + 6d = 14   — (i)

S17 = [2a + 16d] =  289   → 2a + 16d = 34  —— (ii)

Subtracting eq (i) from (ii) we get 10d = 20

d = 2

Putting value of  d in eq (i) we get

2a + 6(2) = 14

a = 1

Now,

Sn = [2a + (n – 1)d]
= [2×1 + (n – 1)2] = [2+ 2n – 2] = [2n] = n2

Q2. If the sum of first 6 terms is 12 and sum of first 10 terms is 60, then find the sum of its n terms.

Ans.  Given S6 = 12 and S10 = 60, Sn = ?

Using Sn = [2a + (n – 1)d]

S6 = [2a + 5d] = 12  → 2a + 5d = 4   — (i)

S10 = [2a + 9d] = 60   → 2a + 9d = 12  —— (ii)

Subtracting eq (i) from (ii) we get 4d = 8

d = 2

Putting value of  d in eq (i) we get

2a + 5(2) = 4

2a = -6

a = – 3

Now,

Sn = [2a + (n – 1)d]
= [2×(-3) + (n – 1)2]

= [ -6 + 2n – 2]     = [2n – 8]

Q3. If the sum of first 10 terms of an A. P. is -60 and sum of first 15 terms is-165, then find the sum of its n terms.

Ans. Given S10 = -60 and S15 = -165, Sn = ?

Using Sn = [2a + (n – 1)d]

S10 = [2a + 9d] = -60   → 2a + 9d = -12   — (i)

S15 = [2a + 14d] = -165   → 2a + 14d = -22  —— (ii)

Subtracting eq (i) from (ii) we get 5d = -10

d = -2

Putting value of d in eq (i) we get

2a + 9(-2) = -12

2a = 6

a =  3

Now,

Sn = [2a + (n – 1)d]
= [2×3 + (n – 1)(-2)]

= [6 – 2n + 2]     = [8 – 2n]

Q4. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference:

Ans. a = 5, l = 45, Sn = 400,   n = ?, d = ?

Using,  Sn = (a + l)

400 = (5 + 45)

= 8

n = 16

Now, Sn = [2a + (n – 1)d]

400 = [2×5 + 15d]

50 = 10 + 15d

40 = 15d

d = =

Q5. How many terms of A. P. : 9, 17, 25, ………. Should be taken so that their sum is 636 ?

Ans. a = 9, d = 17 – 9 = 8, Sn = 636, n = ?

Using, Sn = [2a + (n – 1)d]

636 = [2×9 + (n – 1)8]

636 = [8n + 10]

636 = 4n2 + 5n

4n2 + 5n – 636 = 0

4n2 + 53n – 48n – 636 = 0

n(4n – 53) – 12(4n – 53) = 0

(n – 12)(4n – 53) = 0

n = 12 or n =

As n cannot be in fraction. Therefore, n = 12 is correct.

Q6. Which term of A.P. 3, 15, 27, 39………… will be 132 more than its 54th term?

Ans. a = 3, d = 15 – 3 = 12

an = a + (n – 1)d

a54 = 3 + (54 – 1)12 = 3 + 53×12 = 639

According to question,

We have to find the term which is equal to a54 + 132 i.e 639 + 132 = 771

an = a + (n – 1)d

771 = 3 + (n – 1)12

768 = (n – 1)12

64 = n – 1

65 = n

Therefore, 65th term will be 132 more than 54th term of given A.P.

Q7. Find the 20th term from the last of the A. P. 3, 8, 13, …………, 253.

Ans. Now AP will be 253, 248, ……. 3

a = 253, d = – 5, a20 = ?

an = a + (n – 1)d

a20 = 253 + 19(-5) = 253 – 95 = 158

Q8. Find the 31st term of an A. P. whose 11th term is 38 and 16th term is 73.

Ans. a31 = ?, a11 = 38, a16 = 73

an = a + (n – 1)d

a11 = a + 10d = 38   — (i)

a16 = a + 15d = 73  — (ii)

Subtracting eq (i) from (ii) we get 5d = 35

d = 7

Putting value of d in equation (i)

a + 10×7 = 38

a = – 32

Now, an = a + (n – 1)d

a31 = – 32 + 30×7 = – 32 + 210 = 178

Q9. The sum of first n natural numbers is __________.

Ans. Natural numbers form an AP : 1, 2, 3, …… n

a = 1, d = 1, Sn = ?

Using, Sn = [2a + (n – 1)d]

Sn = [2×1 + (n – 1)1] = [n + 1]

Q10. The sum of first 50 natural numbers is ________ .

Ans. Natural numbers form an AP : 1, 2, 3, ……

a = 1, d = 1, S50 = ?

Using, Sn = [2a + (n – 1)d]

S50 = [2×1 + (50 – 1)1] = 25(51) = 1275

Q11. Write the next four terms of the A. P. 1, -1, -3, -5, ……………\

Ans. a = 1, d = -1 – 1 = -2,

Next term comes when we add d in previous term.

-5 + (-2) = – 7

-7 + (-2) = -9

-9 + (-2) = – 11

-11 + (-2) = -13

Therefore, required next four terms are : -7, -9, -11, -13

Q12. Find the sum of first 10 terms of A.P. 2, 7, 12, 17, …………..

Ans. a = 2, d = 7 – 2 = 5, S10 = ?

Using, Sn = [2a + (n – 1)d]

Sn = [2×2 + (10 – 1)5] = 5[4 + 45] = 5×49 = 245

Q13. Find 11th term of A.P. 7, 13, 19 …………. Most Important

Ans. a = 7, d = 13 – 7 = 6, a11 = ?

an = a + (n – 1)d

a11 = a + 10d =  7 + 10×6 = 67

Q14. Find the common difference of the A.P. : 3, 1, −1, -3, ……………. Most Important

Ans. d = a2 – a1 = 1 – 3 = – 2

Q15. Find common difference of AP. 7, 5, 3, 1 ……………

Ans. d = 5 – 7 = – 2

Q16. How many three digit numbers are divisible by 7 ?

Ans. Three digit numbers divisible by 7 form an AP : 105, 112 …….. 994

a = 105, d = 7, an = 994, n = ?

an = a + (n – 1)d

994 = 105 + (n – 1)7

889 = (n – 1)7

127 = (n – 1)

n = 128

Q17. How many two-digit numbers are divisible by 3?

Ans. Two digit number divisible by 3 form an AP : 12, 15, ……. 99

a = 12, d = 3, an = 99, n = ?

an = a + (n – 1)d

99 = 12 + (n – 1)3

87 = (n – 1)3

29 = (n – 1)

n = 30

Q18. Find the sum of first 40 positive integers divisible by 6.

Ans. First 40 positive integers divisible by 6 form an AP : 6, 12, ….

a = 6, d = 6, S40 = ?

Sn = [2a + (n – 1)d]

S40 = [2×6 + (40 – 1)6] = 20[12 + 234]

= 20 × 246 = 4920

Q19. How many multiples of 4 lie between 10 and 250 ?

Ans. Multiples of 4 will form an AP : 12, 16, …… 248

a = 12, d = 4, an = 248

an = a + (n – 1)d

248 = 12 + (n – 1)4

236 = (n – 1)4

59  =  (n – 1)

n = 60

Q20. How many terms of sequence 1, 4, 7, 10, ………. should be taken so that their sum is 176 ?

Ans. Given Sn = 176, a = 1, d = 4 – 1= 3

n = ?

Using,  Sn = [2a + (n – 1)d]

176 = [2 + (n – 1)3]

352 = n[3n – 1]

3n2 – n – 352 = 0         (We have to find two numbers whose sum is – 1 and product is – (3×352) )

3n2 – 33n + 32n – 352 = 0

3n(n – 11) + 32(n – 11) = 0

(3n + 32)(n – 11) = 0

As n cannot be negative.

Therefore, n – 11= 0

n = 11.

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