Solve the equation

Question:

Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots . . \mathrm{a}_{49}$ be in A.P. such that $\sum_{\mathrm{k}=0} \mathrm{a}_{4 \mathrm{k}+1}=416$ and $\mathrm{a}_{9}+\mathrm{a}_{43}=66$. If $\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots \ldots+\mathrm{a}_{17}^{2}$ $=140 \mathrm{~m}$, then $\mathrm{m}$ is equal to-

  1. 68

  2. 34

  3. 33

  4. 66


Correct Option: , 2

Solution:

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