Question

Find the first three terms of each arithmetic series described.

$n=8,a_n=36,S_n=120$

Short answer

The first three terms are-6, 0, 6.

Step-by-step solution

Step 1. Given Information.

Given arithmetic series is $n=8,a_n=36,S_n=120$.

Step 2. Calculation.

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

$S_n=\frac{n}{2}(a_1+a_n)$

Here, $n=8,a_n=36,S_n=120$

Plugging the values:

$\begin{array}{l}{S}_n=\frac{n}{2}(a_1+a_n)\\ 120=\frac{8}{2}(a_1+36)\\ 120=4(a_1+36)\\ a_1+36=30\\ a_1=30-36\\ a_1=-6\end{array}$

The nth term of an arithmetic series is given by

$a_n=a_1+(n-1)d$

Here $a_1=-6,a_n=36,n=8$$a_1=-6,a_n=36,n=8$

Plugging the values:

$\begin{array}{l}{a}_n=a_1+(n-1)d\\ 36=-6+(8-1)d\\ 7d=36+6\\ 7d=42\\ d=6\end{array}$

Adding the common difference d to a term gives the next term.

Hence,

The second term of the series is $-6+6=0$

The third term of the series is $0+6=6$

Step 3. Conclusion.

Hence, the first three terms are -6, 0, 6.