Question

Find the sum of first $31$ term term of an A.P whose $n^{th}$ term is $(3+\frac{2n}{3})$.

• A. $423\frac{2}{3}$
• B. $413\frac{1}{3}$
• C. $417\frac{2}{3}$
• D. $419\frac{2}{3}$

Answer 1

Answer: (A)

$423\frac{2}{3}$

As we know that the sum of first $n$ terms, $S_n=\frac{n}{2}(a+l)$, where $a$ & $l$ are the first term & last term of an AP.

$a_1=3+\frac{2}{3}=\frac{11}{3}$
$a_{31=}=3+\frac{2}{3}\times 31=3+\frac{62}{3}=71$
$\therefore S_{31}=\frac{31}{2}\times (\frac{11}{3}+\frac{71}{3})=\frac{31}{2}\times \frac{82}{3}=\frac{1271}{3}$
$=423\frac{2}{3}$