Consider an A.P. of positive integers, whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$. Then its $11^{\text{th}}$ term is:
Step-by-Step Solution
Key Concept: $S_{3}=3a_{2}=54\Rightarrow a+d=18.$ Substituting into $S_{20}$ inequality gives a window for $d$. Positive-integer A.P.\ forces $d\in\mathbb{N}$, yielding a unique $d$.
$S_{3}=3a+3d=54\Rightarrow a=18-d.$
$S_{20}=10(2a+19d)=10\bigl[2(18-d)+19d\bigr]=10(36+17d).$
$1600<10(36+17d)<1800\Rightarrow 124<17d<144\Rightarrow d\in(7.29,\,8.47).$
Since $d\in\mathbb{N}$, $d=8$ and $a=10$.
$T_{11}=a+10d=10+80=90.$
Correct Answer: 1