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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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