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If a1, a2, a3, …..., an are in A. P. and a1 + a4 + a7 + ...... + a16 = 114, then a1 + a6 + a11 + a16 is equal to:
1. 98
2. 76
3. 38
4. 64

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Correct Answer - Option 2 : 76

Given, a1 + a4 + a7 +…+ a16 = 114

Here, d = 3

⇒ a1 + a4 + a7 + a10 + a13 + a16 = 144

∵ a1 + a16 = a4 + a13 = a7 + a10

⇒ a1 + a16 + a1 + a16 + a1 + a16 = 144

⇒ 3(a1 + a16) = 114

∴ a1 + a16 = 38     ----(1)

Now,

a1 + a6 + a11 + a16

∵ a1 + a16 = a6 + a11

⇒ a1 + a6 + a11 + a16 = a1 + a16 + a1 + a16

⇒ a1 + a6 + a11 + a16 = 2(a1 + a16)

On substituting equation (1),

⇒ 2(a1 + a16) = 2 × 38

∴ a1 + a6 + a11 + a16 = 76

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