# Find the sum of the first n terms of the arithmetic sequence.−9 + (−12) + (−15)+ ... (to 10 terms)

Find the sum of the first n terms of the arithmetic sequence.$-9+\left(-12\right)+\left(-15\right)+...$ (to 10 terms) Find the sum of the first 100 terms of an arithmetic sequence with 15th term of 86 and first term of 2.

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Step 1 Given: $-9+\left(-12\right)+\left(-15\right)+....$ It is an arithmetic series with $a=\text{first term}=-9$
$f=\text{common difference}=-12-\left(-9\right)=-3$ Now sum of the first n terms is, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
$⇒{S}_{n}=n/2\left[2\left(-9\right)+\left(n-1\right)\left(-3\right)\right]$
$⇒{S}_{n}=\frac{n}{2}\left[-18-3n+3\right]$
$⇒{S}_{n}=\frac{n}{2}\left[-15-3n\right]$
$⇒{S}_{n}=\frac{3n}{2}\left[5+n\right]$

Step 2 Given Consider, ${a}_{15}=86$
$⇒a+\left(15-1\right)d=86\left[\because {a}_{n}=a+\left(n-1\right)d\right]$
$⇒2+14d=86$
$⇒14d=84$
$⇒d=6$ Therefore the sum of the first 100 terms is, ${S}_{100}=\frac{100}{2}\left[2\left(2\right)+\left(100-1\right)\left(6\right)\right]\left[\because {S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right]$
$=50\left(4+594\right)$
$=29900$