Question

find the first three terms of the arithmetic series. remember to use the formula $s_{n}=\frac{n}{2}(a_{1}+a_{n})$ to find n, and the formula $a_{n}=a_{1}+(n - 1)d$ to find d where $a_{1}=-10$, $a_{n}=26$, $s_{n}=80$. \
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\bigcirc a. $-10 + 16 + 42 + \dots$ \
\
\bigcirc b. $-10 + -14 + -18 + \dots$ \
\
\bigcirc c. $-10 + -6 + -2 + \dots$ \
\
\bigcirc d. $26 + 22 + 18 + \dots$

Explanation:

Step1: Find number of terms $n$

Substitute $S_n=80$, $a_1=-10$, $a_n=26$ into $S_n=\frac{n}{2}(a_1+a_n)$:
$$80=\frac{n}{2}(-10+26)$$
Simplify the right-hand side:
$$80=\frac{n}{2}(16)$$
$$80=8n$$
Solve for $n$:
$$n=\frac{80}{8}=10$$

Step2: Find common difference $d$

Substitute $a_n=26$, $a_1=-10$, $n=10$ into $a_n=a_1+(n-1)d$:
$$26=-10+(10-1)d$$
Simplify and solve for $d$:
$$26+10=9d$$
$$36=9d$$
$$d=\frac{36}{9}=4$$

Step3: Calculate 2nd and 3rd terms

2nd term: $a_2=a_1+d=-10+4=-6$
3rd term: $a_3=a_2+d=-6+4=-2$

Answer:

c. $-10 + -6 + -2 + ...$