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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:

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answer is C.

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Detailed Solution

Given,
a1 + a4 + a7, a10, a13, a16 = 114
Formula for sum of an A.P.
S =

\frac{n}{2}

⋅ (a1 + an)
Where, an = the last term in the sequence
              a1 = the first term in the sequence
              n = the number of terms in the sequence
              S= Sum of the series
a1 + a4 + a7, a10, a13, a16 = 114

∴

S =

\frac{n}{2}

⋅ (a1 + an)
114 =

\frac{6}{2}

⋅ (a1 + a16)
(a1 + a16) =

\frac{2}{6}

⋅ 114
(a1 + a16) = 38
We have a1 + a6 + a11, a16

∴

S =

\frac{n}{2}

⋅ (a1 + an)
S =

\frac{4}{2}

⋅ (a1 + a16) As, (a1 + a16) = 38
S =

\frac{4}{2}

⋅ (38)
S = 76

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