Question

The sum of $1$st, $3$rd and $17$th term of AP is $216.$ find the sum of the first $13$ terms of the AP.

Answer 1

We know that $n^{th}$ term of an A.P whose first term is $"a"$ and common difference is $"d"$ is given by, $a_n=a+(n-1)d$

So, $1$st term $=a$, $3$rd term $=a+2d$, and $17$th term $=a+16d$

According to question, we have

$1$st term $+3$rd term $+17$th term $=216$

$\Rightarrow a+(a+2d)+(a+16d)=216$

$\Rightarrow 3a+18d=216$

$\Rightarrow 3(a+6d)=216$

$\Rightarrow a+6d=\frac{216}{3}$

$\Rightarrow a+6d=72\dots \dots \left(i\right)$

Sum of $"n"$ terms in an AP, $S_n=\frac{n}{2}\times \left[\right(2a+(n-1)d\left)\right]$

Sum of thirteen terms, $S_{13}=\frac{13}{2}\times \left[\right(2a+12d\left)\right]$

$=13(a+6d)$

$=13\times 72=936$ [Using $eq.\left(i\right)$]