Let the first term and common difference of an AP of positive integers be such that the sum of the first 3 terms is 54 and $1600 < S_{20} < 1800$. Then the $11^{\text{th}}$ term equals
Step-by-Step Solution
Key Concept: Use $S_3=3(a+d)=54$ to get $a+d=18$, express $S_{20}$ in terms of $d$ alone, and apply the strict inequality to find the unique integer $d$.
$S_3=3(a+d)=54\Rightarrow a+d=18$.
$S_{20}=10(2a+19d)=10\bigl(36-2d+19d\bigr)=10(36+17d)=360+170d$.
$1600<360+170d<1800\Rightarrow\frac{1240}{170}<d<\frac{1440}{170}\Rightarrow7.29<d<8.47$.
Since $d$ is a positive integer, $d=8$, $a=10$.
$a_{11}=10+10(8)=90$.
Correct Answer: 1