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Question:

If a_1, a_2, a_3,....a_n are in A.P. and a_1+ a_4 + a_7 + ..... + a_16 = 114, then a_1 + a_6 + a_11 + a_16 is equal to:

38

98

64

76

Solution:

Correct option is C. 76
a1+a4+a7+a10+a13+a16=114⇒6/2(a1+a16)=114⇒a1+a16=38
So, a1+a6+a11+a16=4/2(a1+a16)=2(38)=76