NCERT Solutions for Class 10 Maths Chapter 5

Important NCERT Questions

Arithmetic Progression

NCERT Books for Session 2022-2023

CBSE Board and UP Board Others state Board

EXERCISE 5.3

Page No:107

Questions No:4

# How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

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Here, a = 9, d = 17 – 9 = 8 and S_n = 636.

The sum of n terms of an AP is given by

S_n = n/2[2a + (n -1)d]

⇒ 636 = n/2[2(9) + (n -1)(8)]

⇒ 636 = n[9 + 4n – 4]

⇒ 4n² + 5n – 636 = 0

⇒ 4n² + 53n – 48n – 636 = 0

⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (n – 12)(4n + 53) = 0

⇒ n – 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ – 53/4]

⇒ n = 12

Hence, 12 terms of the AP: 9, 17, 25 … must be taken to get the sum 636.