In an increasing geometric progression of positive terms, the sum of the second and sixth terms is (70)/(3) and the product of the third and fifth terms is 49 . Then the sum of the 4^ th , 6^ th and 8^ th terms is equal

Correct answer: B. 91 T_2+T_6=(70)/(3) ar+ar^5=(70)/(3) ~T_3 · T_5=49 ar^2 · ar^4=49 a^2 r^6=49 ar^3=+7, a= 7 r^3 ar(1+r^4)=(70)/(3) 7 r^2 (1+r^4)=(70)/(3), r^2=t 1 t (1+t^2)=(10)/(3) 3 t^2-10 t+3=0 t=3, (1)/(3) Increasing G.P. r^2=3, r= 3 T_4+T_6+T_8 =ar^3+ar^5+ar^7 =ar^3(1+r^2+r^4) =7(1+3+9)=91

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Sequences and Series

In an increasing geometric progression of positive terms, the sum...

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is

\frac{70}{3}

and the product of the third and fifth terms is 49 . Then the sum of the

4^{\text{th }}, 6^{\text{th }}

and

8^{\text{th }}

terms is equal to :

Options

Correct answer

B. 91

Step-by-step solution

\mathrm{T}_2+\mathrm{T}_6=\frac{70}{3}

\mathrm{ar}+\mathrm{ar}^5=\frac{70}{3}

\mathrm{~T}_3 \cdot \mathrm{T}_5=49

\mathrm{ar}^2 \cdot \mathrm{ar}^4=49

\mathrm{a}^2 \mathrm{r}^6=49

\mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3}

\mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3}

\frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t}

\frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3}

3 \mathrm{t}^2-10 \mathrm{t}+3=0

\mathrm{t}=3, \frac{1}{3}

Increasing G.P.

\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}

\mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8

=\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7

=\operatorname{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right)

=7(1+3+9)=91

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