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

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

Sequences and Series

Solution:

$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\times10$