How many terms of the series -20, -16, -12, ... must be

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How many terms of the series -20, -16, -12, ... must be taken so that the sum is 120 ?

1. 5
2. 20
3. 15
4. 10

Correct Option: C

S_n = 120, a = -20, d = 4
S_n = n/2[2a + (n - 1)d ]
⇒ 120 = n/2 [2 x (-20) + (n - 1) x 4]
⇒ 120 = -20n + 2(n -1)n
⇒ 120 = -20n + 2n² - 2n
⇒ 120 = 2n² - 22n
⇒ 2n² - 22n - 120 = 0
⇒ n²-11n - 60 = 0
⇒ n² - 15n + 4n - 60 = 0
⇒ n(n - 15) + 4(n - 15) = 0
⇒ (n + 4) (n - 15) = 0
⇒ n = -4 or 15
∴ n = 15