Verified Solution • 2025–26 Session moderate QJEEM24Ma7a46: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7a46 Solution: Let , ............ be an infinite G.P. If ∑ n = 0 ∞ a r n = 57… Let a,ar,ar2a, a r, a r^2a,ar,ar2 , ............ be an infinite G.P. If ∑n=0∞arn=57 and ∑n=0∞a3r3n=9747, then a+18ra+18 ra+18r is equal to Multiple Choice Options A 27 B 38 C 31 D 46 Step-by-Step Verified Solution ∑n=0∞arn=57amp;⇒a1−r=57.... (i)∑n=0∞a3r3n=9747amp;⇒a31−r3=9747.... (ii)\begin{array}{ll}\sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\\sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)}\end{array}∑n=0∞arn=57∑n=0∞a3r3n=9747amp;⇒1−ra=57.... (i)amp;⇒1−r3a3=9747.... (ii) amp;(1−r3)(1−r)3=(57)39747=19amp;⇒(1−r)(1+r+r2)(1−r)3=19amp;⇒18r2−39r+18=0amp;⇒r=23,32 (rejected) amp;∴a=19amp;a+18ramp;=19+12=31\begin{aligned}& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\& \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\& \Rightarrow \quad 18 r^2-39 r+18=0 \\& \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\& \therefore \quad a=19 \\& \quad a+18 r \\& \quad=19+12=31\end{aligned}amp;(1−r)3(1−r3)=9747(57)3=19amp;⇒(1−r)3(1−r)(1+r+r2)=19amp;⇒18r2−39r+18=0amp;⇒r=32,23 (rejected) amp;∴a=19amp;a+18ramp;=19+12=31 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