Question

The first term of an $A.P.$ is $10$ and its ${41}^{st}$ term is $81$. Find the sum of the series, if $81$ is the last term.

• A. $1025$
• B. $1865.5$
• C. $858$
• D. $962$

Answer 1

The correct option is B $1865.5$

Given, first term, $a=10$

last term, $l=81$

total number of terms, $n=41$

Sum, $S_n=\frac{n}{2}[a+l]$

$=\frac{41}{2}[10+81]$

$=\frac{41}{2}\times 91$

$=1865.5$

Hence, the sum of the series is $1865.5$