Question

In an A.P. of 81 terms, the 41th term is 10. Then the sum of the series is

• A. $14\times 81$
• B. $10\times 81$
• C. $10\times 41$
• D. $\frac{10\times 41}{2}$

Answer 1

Answer: (B)

$T_{41}=a+40d=10$
$S_{81}=\frac{81}{2}\left[2a+\left(81-1\right)d\right]$
$=\frac{81}{2}\left[2a+80d\right]=81\left[a+40d\right]$
$=81\times 10$