Find $a_{1} + a_{6} + a_{11} + a_{16}$ if it is known that $a_{1}, a_{2}$ , ___________ is an A.P. and $a_{1} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16} = 147$ .

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Solution

We know that in AP the sum of first and last term is equal to second and second last term and so

$a_{1} + a_{4} + a_{7} + a_{10} + a_{13} + a_{16} = 147$

$(a_{1} + a_{16}) + (a_{4} + a_{13}) + (a_{7} + a_{10}) = 147$

$(a_{1} + a_{16}) = (a_{4} + a_{13}) = (a_{7} + a_{10}) = \frac{147}{3} = 49$

So asked values,

$a_{1} + a_{16} + a_{11} + a_{6} = (a_{1} + a_{16}) + (a_{11} + a_{6}) = 2 (a_{1} + a_{16}) = 2 \times 49 = 98$

Suggest Corrections

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

a_{1}, a_{2} . . . .

a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147.

a_{n}

a_{1} + a_{4} + a_{7} + . . . + a_{16} = 147

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

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}

a_{1}, a_{2}, a_{3}

a_{4} + a_{8} + a_{12} + a_{16} = 224

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