Question

In an A.P of 81 terms the 41st is ten. then find thee sum of tge series ​

Answer 1

$\small\mathcal\color{green}QueStiOn$:

In an A.P. of $81 terms$ the ${41}^{st}$ is $10$, then find the sum of the series.

$\huge\underline\mathbb\color{red}ANSWER$ :

${a}_{41} = a + 40d = 10$

${S}_{n} =\frac{81}{2}[2a + (81 - 1) d]$

${S}_{n} = \frac {81}{2}[2a + 80d]$

${S}_{n} = 81 × 10$

${S}_{n} = 810$