Question

How many terms of the AP 3, 5, 7, … to be added to get a sum 120?​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

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

Given,

Sum = 120

a = 3 (first term)

d = 2 (difference)

120 = n/2 [6 + (n - 1) 2]

n/2 ( 6 + 2n - 2) = 120

Get 2 common in LHS

n (3 + n - 1) = 120

n (n + 2) = 120

n^2 + 2n = 120

n^2 + 2n - 120 = 0

n^2 - 10n + 12n - 120 = 0

n (n - 10) + 12 (n - 10) =0

(n + 12) (n - 10) = 0

n = -12, 10

Since the no. of terms cannot be negative, therefore

n = 10

No. of terms = 10