Question
Find the sum of $-5 + (-8) + (-11) + …. + (-230).$

Answer

Let $S_n = –5 + (–8) + (–11) + …. + (–230)$
Clearly, the terms of the sum form an A.P.
with, $a = –5$
$d = –8 –(–5) = –8 + 5 = –3$
$l = –230$
Let the number of terms of the AP be n
We know that
$l = a + (n - 1)d$
$ \Rightarrow  -230 = -5 + (n - 1) (-3)$
$ \Rightarrow  (n - 1) (-3) = -230 + 5$
$ \Rightarrow  (n - 1) (-3) = -225$
$ \Rightarrow n - 1 = \frac{{ - 225}}{{ - 3}} = 75$
$ \Rightarrow  n = 75 + 1$
$ \Rightarrow  n = 76$
Again, we know that
${S_n} = \frac{n}{2}(a + l)$
$ \Rightarrow {S_{76}} = \frac{{76}}{2}\left[ {( - 5) + ( - 230)} \right]$
$ \Rightarrow  S_{76} = 38(-235)$
$ \Rightarrow  S_{76} = -8930$
Hence, the required sum is $-8930.$

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