If $a_1, a_2, a_3, \ldots, a_n, \ldots$. are in A.P. such that $a_4-a_7$ $+a_{10}=m$, then the sum of first…
If $a_1, a_2, a_3, \ldots, a_n, \ldots$. are in A.P. such that $a_4-a_7$ $+a_{10}=m$, then the sum of first 13 terms of this A.P., is :
$10 \mathrm{~m}$
$12 \mathrm{~m}$
$13 \mathrm{~m}$
$15 \mathrm{~m}$
Solution
If $d$ be the common difference, then
$
\begin{aligned}
m & =a_4-a_7+a_{10}=a_4-a_7+a_7+3 \mathrm{~d}=a_7 \\
\mathrm{~S}_{13} & =\frac{13}{2}\left[a_1+a_{13}\right]=\frac{13}{2}\left[a_1+a_7+6 d\right] \\
& =\frac{13}{2}\left[2 a_7\right]=13 a_7=13 \mathrm{~m}
\end{aligned}
$