Let a_1, a_2, a_3, be a G. P. of increasing positive numbers. If a_3 a_5=729 and a_2+a_4=(111)/(4) , then 24 (a_1+a_2+a_3) is equal to

Correct answer: C. 129 Let the I^ st term of G.P. be a \& common ratio be r aligned a_3 a_5 & =ar^2 · ar^4=729 & =a^2 r^6=729 & =ar^3=27 ....(i) aligned aligned a_2+a_4 & =ar+ar^3=(111)/(4) & =ar=(3)/(4) ....(ii) aligned (i) (ii) aligned & ar^3 ar =(27)/(3 / 4) & r^2=36 & r=6 aligned from (ii) a(6)=(3)/(4) a=(1)/(8) Now, 24 (a_1+a_2+a_3) aligned & =24(a+a r+a^2) & =24 a(1+r+r^2) aligned aligned & =24 × (1)/(8)(1+6+36) & =3(43) & =129 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Sequences and Series

Let a_1, a_2, a_3, be a G. P. of increasing positive numbers. If...

Let

a_1, a_2, a_3, \ldots be a G. P. of increasing positive numbers. If

\mathrm{a}_3 \mathrm{a}_5=729

and

\mathrm{a}_2+\mathrm{a}_4=\frac{111}{4}

, then

\left(a_1+a_2+a_3\right)

is equal to

Options

A $131$
B $130$
C $129$
D $128$

Correct answer

C. $129$

Step-by-step solution

Let the

\mathrm{I}^{\text{st }}

term of G.P. be a \& common ratio be r

\begin{aligned}<br/>\mathrm{a}_3 \mathrm{a}_5 & =\operatorname{ar}^2 \cdot \operatorname{ar}^4=729 \\ & =\mathrm{a}^2 \mathrm{r}^6=729 \\ & =\mathrm{ar}^3=27 \quad ....(i)<br/>\end{aligned}

\begin{aligned} \mathrm{a}_2+\mathrm{a}_4 & =\mathrm{ar}+\mathrm{ar}^3=\frac{111}{4} \\ & =\mathrm{ar}=\frac{3}{4} \quad ....(ii)\end{aligned}

\text{ (i) } \div \text{ (ii) }

\begin{aligned}<br/>& \frac{\mathrm{ar}^3}{\mathrm{ar}}=\frac{27}{3 / 4} \\ & \mathrm{r}^2=36 \\ & \mathrm{r}=6<br/>\end{aligned}

from (ii)

a(6)=\frac{3}{4} \Rightarrow a=\frac{1}{8}

Now,

24

\left(a_1+a_2+a_3\right)

\begin{aligned}<br/>& =24\left(a+a r+a^2\right) \\ & =24 a\left(1+r+r^2\right)<br/>\end{aligned}

\begin{aligned} & =24 \times \frac{1}{8}(1+6+36) \\ & =3(43) \\ & =129\end{aligned}

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