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

Class	10
Chapter	Arithmetic Progressions
Subject	Math
Category	Important Question Answer

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Class 10 Math Chapter 5 Important Question Answer

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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, S_n = ?

Using S_n = [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,

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

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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, S_n = ?

Using S_n = [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,

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

= [ -6 + 2n – 2]     = [2n – 8]

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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, S_n = ?

Using S_n = [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,

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

= [6 – 2n + 2]     = [8 – 2n]

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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, S_n = 400,   n = ?, d = ?

Using,  S_n = (a + l)

400 = (5 + 45)

= 8

n = 16

Now, S_n = [2a + (n – 1)d]

400 = [2×5 + 15d]

50 = 10 + 15d

40 = 15d

d = =

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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, S_n = 636, n = ?

Using, S_n = [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.

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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

a_n = 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

a_n = a + (n – 1)d

771 = 3 + (n – 1)12

768 = (n – 1)12

64 = n – 1

65 = n

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

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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 = ?

a_n = a + (n – 1)d

a₂₀ = 253 + 19(-5) = 253 – 95 = 158

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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

a_n = 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, a_n = a + (n – 1)d

a₃₁ = – 32 + 30×7 = – 32 + 210 = 178

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

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

a = 1, d = 1, S_n = ?

Using, S_n = [2a + (n – 1)d]

S_n = [2×1 + (n – 1)1] = [n + 1]

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

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

a = 1, d = 1, S₅₀ = ?

Using, S_n = [2a + (n – 1)d]

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

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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

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

Ans. a = 2, d = 7 – 2 = 5, S₁₀ = ?

Using, S_n = [2a + (n – 1)d]

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

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

Ans. a = 7, d = 13 – 7 = 6, a₁₁ = ?

a_n = a + (n – 1)d

a₁₁ = a + 10d =  7 + 10×6 = 67

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

Ans. d = a₂ – a₁ = 1 – 3 = – 2

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

Ans. d = 5 – 7 = – 2

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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, a_n = 994, n = ?

a_n = a + (n – 1)d

994 = 105 + (n – 1)7

889 = (n – 1)7

127 = (n – 1)

n = 128

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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, a_n = 99, n = ?

a_n = a + (n – 1)d

99 = 12 + (n – 1)3

87 = (n – 1)3

29 = (n – 1)

n = 30

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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 = ?

S_n = [2a + (n – 1)d]

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

= 20 × 246 = 4920

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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, a_n = 248

a_n = a + (n – 1)d

248 = 12 + (n – 1)4

236 = (n – 1)4

59  =  (n – 1)

n = 60

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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,  S_n = [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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