Verified Solution • 2025–26 Session moderate QJEEM24Ma011c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma011c Solution The 20th term from the end of the progression 20,1914,1812,1734,…,−12914 is : \text { The } 20^{\text {th }} \text { term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text { is : } The 20th term from the end of the progression 20,1941,1821,1743,…,−12941 is : Multiple Choice Options A −115-115−115 B −100-100−100 C −110-110−110 D −118-118−118 Step-by-Step Verified Solution 20,1914,1812,1734,……,−1291420,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}20,1941,1821,1743,……,−12941 This is A.P. with common difference amp;d1=−1+14=−34amp;−12914,…………...1914,20\begin{aligned}& d_1=-1+\frac{1}{4}=-\frac{3}{4} \\& -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20\end{aligned}amp;d1=−1+41=−43amp;−12941,…………...1941,20 This is also A.P. a=−12914\mathrm{a}=-129 \frac{1}{4}a=−12941 and d=34\mathrm{d}=\frac{3}{4}d=43 Required term === amp;−12914+(20−1)(34)amp;=−129−14+15−34=−115\begin{aligned}& -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\& =-129-\frac{1}{4}+15-\frac{3}{4}=-115\end{aligned}amp;−12941+(20−1)(43)amp;=−129−41+15−43=−115 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring