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If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}, . . .$ are in A.P. such that $a_{4} - a_{7} + a_{10} = m,$ then sum of the first 13 terms of the A.P. is

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$10 m$

$12 m$

$13 m$

$15 m$

answer is C.

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Detailed Solution

Let d be the common difference of the A.P.,

then $\begin{array}{l} m = a_{4} - a_{7} + a_{10} = a_{4} + (a_{10} - a_{7}) \\ = a_{1} + 3 d + 3 d = a_{1} + 6 d \end{array}$

Now, $S_{13} = \frac{13}{2} [2 a_{1} + (13 - 1) d] = 13 m .$

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