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 11th term is :
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \mathrm{S}_3=3 \mathrm{a}+3 \mathrm{~d}=54 \\
& \Rightarrow \mathrm{a}+\mathrm{d}=18 \\
& \mathrm{~S}_{20}=10(2 \mathrm{a}+19 \mathrm{~d}) \\
& \Rightarrow 10(36+17 \mathrm{~d}) \\
& \Rightarrow 1600<10(36+17 \mathrm{~d})<1800 \\
& \Rightarrow 160<36+17 \mathrm{~d}<180 \\
& \Rightarrow 124<17 \mathrm{~d}<144 \\
& \Rightarrow 7 \frac{5}{17}<\mathrm{d}<8 \frac{8}{17}
\end{aligned}\\
&\text { Common difference will be natural number }\\
&\begin{aligned}
& \Rightarrow d=8 \Rightarrow a=10 \\
& \Rightarrow a_{11}=10+10 \times 8=90
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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