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Sum of n Terms

Let a 1, a 2,...

Question

Let $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ be in A.P. If $a_{3} + a_{7} + a_{11} + a_{15} = 72$ , then the sum of its first $17$ terms is equal to:

204

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612

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153

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306

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Solution

The correct option is D $306$
$a_{3} + a_{7} + a_{11} + a_{15} = 72$ can be written as
$a_{1} + 2 d + a_{1} + 6 d + a_{1} + 10 d + a_{1} + 14 d = 72 4 (a_{1} + 8 d) = 72 \Rightarrow a_{1} + 8 d = 18$
Now sum of $17$ terms of the $A . P$
$S_{17} = \frac{17}{2} (2 a_{1} + 16 d) \Rightarrow S_{17} = 306$

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