Let $S_n$ denote the sum of the first $n$ terms of an AP. If $S_{40}=1030$ and $S_{12}=57$, then $S_{30}-S_{10}$ equals
Step-by-Step Solution
Key Concept: Form two equations from $S_{40}$ and $S_{12}$ in $a$ and $d$, solve the system, then compute $S_{30}-S_{10}$ directly.
$S_{40}=20(2a+39d)=1030\Rightarrow 2a+39d=51.5$.
$S_{12}=6(2a+11d)=57\Rightarrow 2a+11d=9.5$.
Subtracting: $28d=42\Rightarrow d=\frac{3}{2}$. Then $2a=9.5-\frac{33}{2}=-7\Rightarrow a=-\frac{7}{2}$.
$S_{30}=15\left(-7+29\cdot\frac{3}{2}\right)=15(36.5)=547.5$.
$S_{10}=5\left(-7+9\cdot\frac{3}{2}\right)=5(6.5)=32.5$.
$S_{30}-S_{10}=547.5-32.5=515$.
Correct Answer: 3