JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

Let

\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{49}

be in A.P. such that

\sum_{\mathrm{k}=0}^{12} \mathrm{a}_{4 \mathrm{k}+1}=416

and

\mathrm{a}_{9}+\mathrm{a}_{43}=66

. If

\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots+\mathrm{a}_{17}^{2}=140 \mathrm{m}

, then

\mathrm{m}

is equal to :

View answer and explanation

Correct option: B

The correct answer is 34 because after finding the values of

a_1

and

d

, and substituting them into the given summation, we can solve for

m

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