Given:
[tex](-59)+(-56)+(-53)+\text{.}\ldots\ldots\text{.}[/tex]The given series is in arithmetic progression with a common difference of 3.
The sum of the first n terms in the arithmetic progression is given as,
[tex]\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ a=-59,d=3 \\ S_{40}=\frac{40}{2}\lbrack2(-59)+(40-1)3\rbrack \\ =20\lbrack-118+117\rbrack \\ =-20 \end{gathered}[/tex]Answer: the sum of the first 40 terms in the given series is -20.
