Question

Sum $a-3b,2a-5b,3a-7b$, .... to $40$ terms.

Answer 1

Sum of the series $(a-3b)+(2a-5b)+........40$ terms

The $n^{th}$ term of the series $=T_n$ $=na-(2n+1)b$

Sum $=\sum _{n=1}^{40}(na-(2n+1\left)b\right)$

$=\sum _{n=1}^{40}na-\sum _{n=1}^{40}(2n+1)b$

$=a\sum _{n=1}^{40}n-2b\sum _{n=1}^{40}n-b\sum _{n=1}^{40}1$

$=(a-2b)\times \frac{40(40+1)}{2}-b\left(40\right)$

$=820a-1640b-40b$

$=820a-1680b$