QJEEM24Ma1255: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1255 Solution: Let be the sum to -terms of an arithmetic progression , If 40 < …

$\begin{aligned}&amp; \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\&amp;=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\&amp;=2 \mathrm{n}^2+\mathrm{n} \\&amp; \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K}^2+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K} \\&amp;=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\&amp;=\mathrm{n}(\mathrm{n}+1)\left[\frac{2 \mathrm{n}+1}{3}+\frac{1}{2}\right] \\&amp;=\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}+5)}{6} \\&amp; \Rightarrow 40&lt;\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{k=1}^n S_{\mathrm{k}}&lt;42 \\&amp; 40&lt;4 \mathrm{n}+5&lt;42 \\&amp; 35&lt;4 n&lt;37 \\&amp; \mathrm{n}=9\end{aligned}$