If $a_{n}$ is an A.P and $a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147$ , then $a_{1} + a_{6} + a_{11} + a_{16} =$

96

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98

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100

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None of these

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Solution

The correct option is C $98$
Given equation can be written as
$a_{1} + (a_{1} + 3 d) + (a_{1} + 6 d) + . . . . + a_{1} + 15 d = 147$
$\Rightarrow 6 a_{1} + 3 d (1 + 2 + . . + 5) = 147$
$\Rightarrow 2 a_{1} + 15 d = 49$
Now,
$a_{1} + a_{6} + a_{11} + a_{16} = 4 a_{1} + 5 d + 10 d + 15 d$
$= 4 a_{1} + 30 d = 2 (2 a_{1} + 15 d) = 2 \times 49$
$= 98$
$∴$ Option B is correct.

Suggest Corrections

a_{1}, a_{2}, a_{3}, \dots, a_{n}

a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114

a_{1} + a_{6} + a_{11} + a_{16}

26

169,

a_{1} + a_{6} + a_{11} + a_{16} + a_{21} + a_{26}

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