Question

Find $S_n$ for each arithmetic series described.

localid="1648548029213" $a_1=132,d=-4,a_n=52$

Short answer

$S_n$ for the arithmetic series is$S_{21}=1932$ .

Step-by-step solution

Step 1. Given Information.

Given arithmetic series is $a_1=132,d=-4,a_n=52$.

Step 2. Calculation.

The n^th term of an arithmetic series is given by $a_n=a_1+(n-1)d$

Here, $a_1=132,d=-4,a_n=52$$a_1=132,d=-4,a_n=52$

Plugging the values:

$\begin{array}{l}{a}_n=a_1+(n-1)d\\ 52=132+(n-1)(-4)\\ 52-132=(n-1)(-4)\\ (n-1)(-4)=-80\\ n-1=20\\ n=20+1\\ n=21\end{array}$

The sum $S_n$ of the first n terms of an arithmetic series is given by

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

Here, $a_1=132,n=21,d=-4$$a_1=132,n=21,d=-4$

Plugging the values:

$\begin{array}{l}{S}_n=\frac{n}{2}(2a_1+(n-1)d)\\ S_{21}=\frac{21}{2}(2\left(132\right)+(21-1)(-4))\\ S_{21}=\frac{21}{2}(264+\left(20\right)(-4))\\ S_{21}=\frac{21}{2}(264-80)\\ S_{21}=\frac{21}{2}\left(184\right)\\ S_{21}=21\left(92\right)\\ S_{21}=1932\end{array}$

Step 3. Conclusion.

Hence, the sum is$S_{21}=1932$.