Let a_n denote the nth term of an A.P.

Question

Let $a_n$ denote the $n$th term of an A.P. If the sum of the first $n$ terms, $S_n=700$, $a_6=7$, and $S_7=7$, then $a_n$ is equal to:

Accepted Answer

$S_7 = \dfrac72[2a+6d]=7 \Rightarrow a+3d=1$. Also $a_6=a+5d=7$. Subtracting: $2d=6 \Rightarrow d=3$, so $a=-8$.

$$S_n = \frac{n}{2}\big[2(-8)+(n-1)3\big]=700 \implies n(3n-19)=1400 \implies 3n^2-19n-1400=0.$$

Solving: $n = \dfrac{19\pm131}{6}$, giving $n=25$ (the negative root is rejected).

$a_{25} = -8+24(3) = 64.$

Solutions · 1