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 :

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{S}_3=3 \mathrm{a}+3 \mathrm{~d}=54 \\ & \Rightarrow \mathrm{a}+\mathrm{d}=18 \\ & \mathrm{~S}_{20}=10(2 \mathrm{a}+19 \mathrm{~d}) \\ & \Rightarrow 10(36+17 \mathrm{~d}) \\ & \Rightarrow 1600 < 10(36+17 \mathrm{~d}) < 1800 \\ & \Rightarrow 160 < 36+17 \mathrm{~d} < 180 \\ & \Rightarrow 124 < 17 \mathrm{~d} < 144 \\ & \Rightarrow 7 \frac{5}{17} < \mathrm{d} < 8 \frac{8}{17}
\end{aligned}$
Common difference will be natural number
$\begin{aligned}
& \Rightarrow d=8 \Rightarrow a=10 \\ & \Rightarrow a_{11}=10+10 \times 8=90
\end{aligned}$ /

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first…

Solution