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 $a_{1}=-2$, $a_{n}=-54$, $s_{n}=-392$.
a. $-2 + -6 + -10$
b. $-2 + -56 + -110$
c. $-2 + 2 + 6$
d. $-54 + -50 + -46$

Explanation:

Step1: Solve for $n$ using $S_n$ formula

Substitute $a_1=-2$, $a_n=-54$, $S_n=-392$ into $S_n=\frac{n}{2}(a_1+a_n)$:
$$-392=\frac{n}{2}(-2 + (-54))$$
$$-392=\frac{n}{2}(-56)$$
$$-392=-28n$$
$$n=\frac{-392}{-28}=14$$

Step2: Solve for common difference $d$

Substitute $a_1=-2$, $a_n=-54$, $n=14$ into $a_n=a_1+(n-1)d$:
$$-54=-2+(14-1)d$$
$$-54=-2+13d$$
$$-54+2=13d$$
$$-52=13d$$
$$d=\frac{-52}{13}=-4$$

Step3: Find 2nd and 3rd terms

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

Answer:

a. -2 + -6 + -10