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.

You can still ask an expert for help

Want to know more about Polynomial arithmetic?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

smallq9

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, ${S}_{100}=\frac{100}{2}\left[2\left(2\right)+\left(100-1\right)\left(6\right)\right]\left[\because {S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right]$
$=50\left(4+594\right)$
$=29900$