Question
Find the sum:
$(-5) + (-8) + (-11) + ..... + (-230).$
$(-5) + (-8) + (-11) + ..... + (-230).$
✓
Answer
Given,
$\text { A.P. }(-5)+(-8)+\ldots . . .(-230)$
Here,
First term $a=-5$
Difference $d=-8-(-5)=-8+5=-3$
and Last term $a_n=-230$
We know $a_n=a+(n-1) d$
$\Rightarrow-230=-5+(n-1)(-3)$
$\Rightarrow-230=-5-3 n+3$
$\Rightarrow-230=-2-3 n$
$\Rightarrow-228=-3 n$
$\Rightarrow n=\frac{-228}{-3}=76$
We know, sum of $n$ terms,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{76}=\frac{76}{2}[2(-5)+(76-1)(-3)]$
$\Rightarrow S_{76}=38[-10+75 \times-3]$
$\Rightarrow S_{76}=38[-10-225]$
$\Rightarrow S_{76}=38[-235]$
$\Rightarrow S_{76}=-8930$
Hence, Sum of given A.P. is $-8930$.
$\text { A.P. }(-5)+(-8)+\ldots . . .(-230)$
Here,
First term $a=-5$
Difference $d=-8-(-5)=-8+5=-3$
and Last term $a_n=-230$
We know $a_n=a+(n-1) d$
$\Rightarrow-230=-5+(n-1)(-3)$
$\Rightarrow-230=-5-3 n+3$
$\Rightarrow-230=-2-3 n$
$\Rightarrow-228=-3 n$
$\Rightarrow n=\frac{-228}{-3}=76$
We know, sum of $n$ terms,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{76}=\frac{76}{2}[2(-5)+(76-1)(-3)]$
$\Rightarrow S_{76}=38[-10+75 \times-3]$
$\Rightarrow S_{76}=38[-10-225]$
$\Rightarrow S_{76}=38[-235]$
$\Rightarrow S_{76}=-8930$
Hence, Sum of given A.P. is $-8930$.
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