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How many terms of A.P. 3, 5, 7, 9, ..... must be taken to get the sum 120?

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Question

How many terms of A.P. 3, 5, 7, 9, ..... must be taken to get the sum 120?

Sum
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Solution

Given, A.P. : 3, 5, 7, 9, ..................

First term, a = 3

Common difference, d = 2

Also, given sum, Sn = 120

∴ `S_n = n/2 [2a + (n - 1)d]`

⇒ `120 = n/2 [2 xx 3 + (n - 1)2]`

⇒ 240 = n(2n + 4)

⇒ 120 = n(n + 2)

⇒ n2 + 2n – 120 = 0

⇒ n2 + 12n – 10n – 120 = 0

⇒ n(n + 12) – 10(n + 12) = 0

⇒ (n + 12) (n – 10) = 0

⇒ n = 10 or n = –12   ...[Neglecting because n can’t be negative]

∴ n = 10

Hence, 10 terms must be taken to get the sum 120.

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