Question

value 2
find the first three terms of the arithmetic series. remember to use the form $s_n = \frac{n}{2}(a_1 + a_n)$ to find n, and the formula $a_n = a_1 + (n - 1)d$ to find d where $a_1 = 8$, $a_n = 38$, $s_n = 138$.
\bigcirc a. $8 + 14 + 20 + \dots$
\bigcirc b. $8 + 46 + 84 + \dots$
\bigcirc c. $8 + 2 + -4 + \dots$
\bigcirc d. $36 + 32 + 26 + \dots$

Explanation:

Step1: Solve for $n$ using $S_n$

Substitute $S_n=138$, $a_1=8$, $a_n=38$ into $S_n=\frac{n}{2}(a_1+a_n)$:
$$138=\frac{n}{2}(8+38)$$
Simplify and solve for $n$:
$$138=\frac{n}{2}(46) \implies 138=23n \implies n=\frac{138}{23}=6$$

Step2: Solve for common difference $d$

Substitute $a_n=38$, $a_1=8$, $n=6$ into $a_n=a_1+(n-1)d$:
$$38=8+(6-1)d$$
Simplify and solve for $d$:
$$38-8=5d \implies 30=5d \implies d=\frac{30}{5}=6$$

Step3: Find 2nd and 3rd terms

2nd term: $a_2=a_1+d=8+6=14$
3rd term: $a_3=a_2+d=14+6=20$

Answer:

a. $8 + 14 + 20 + ...$