Verified Solution • 2025–26 Session moderate QJEEM22Madb1f: JEE Main 2022 Mathematics - Class 11 JEEM22Madb1f Solution: Let be an A.P. If , then is equal to _________. Let a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldotsa1,a2,a3,… be an A.P. If ∑r=1∞ar2r=4\sum\limits_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4r=1∑∞2rar=4 , then 4a24 a_{2}4a2 is equal to _________. Step-by-Step Verified Solution Given S=a12+a222+a323+a424 + ..... ∞S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\inftyS=2a1+22a2+23a3+24a4+.....∞ 12S=a122+a223 + ......... ∞S2=a12+(a2+a1)22+(a3+a2)23 + ...... ∞{{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}2S=2a1+22(a2+a1)+23(a3+a2)+......∞21S=22a1+23a2+.........∞ ⇒S2=a12+d2\Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}⇒2S=2a1+2d ⇒a1+d=a2=4⇒4a2=16\Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16⇒a1+d=a2=4⇒4a2=16 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