JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 4)

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 :
108
90
122
84

Explanation

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

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