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Question:

If a_1, a_2, a_3, .... are in A.P. such that a_1 + a_7 + a_16 = 40, then the sum of the first 15 terms of this A.P. is

200

120

280

150

Solution:

Correct option is A. 200
a1+a7+a16=40
a+a+6d+a+15d=40⇒3a+21d=40⇒a+7d=40/3
S15=15/2(2a+14d)=15(a+7d)
S15=15 × 40/3 = 200⇒ S15=200.