Question

The third term of an arithmetic progression is $18$, and the seventh term is $30$, then the sum of $17$ terms is

• A. $612$
• B. $600$
• C. $656$
• D. None of these

Answer 1

The correct option is A $612$

Let $T_n$ be the $n^{th}$ term of the AP.

$T_n=a+(n-1)d$

Where $a$ and $d$ are first term and common difference respectively.

According to the given condition

$T_3=a+(3-1)d$

$T_3=a+2d$

$18=a+2d$........$\left(1\right)$

$T_7=a+(7-1)d$

$T_7=a+6d$

$30=a+6d$........$\left(2\right)$

solving $\left(1\right)$ and $\left(2\right)$ we get,

$a=12$

$d=3$

Sum of $n$ terms of an AP is given as,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{17}=\frac{17}{2}\left[2\right(12)+(17-1\left)3\right]$

$S_{17}=612$

Hence option (A) is correct option