Single Choice

If $$a_1, a_2, a_3,....a_n$$ are in A.P. and $$a_1+ a_4 + a_7 + ..... + a_{16} = 114$$, then $$a_1 + a_6 + a_{11} + a_{16}$$ is equal to:

A$$38$$
B$$98$$
C$$76$$
Correct Answer
D$$64$$

Solution

$$a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114$$
$$\Rightarrow \dfrac{6}{2} (a_1 + a_{16}) = 114$$
$$\Rightarrow a_1 + a_{16} = 38$$
So, $$a_1 + a_6 + a_{11} + a_{16} = \dfrac{4}{2} (a_1 + a_{16})$$
$$=2\times 38 = 76$$

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