Search any question & find its solution
Question:
Answered & Verified by Expert
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:
Solution:
1620 Upvotes
Verified Answer
The correct answer is:
91
$\mathrm{T}_2+\mathrm{T}_6=\frac{70}{3}$
$\mathrm{ar}+\mathrm{ar}^5=\frac{70}{3}$
$\mathrm{~T}_3 \cdot \mathrm{T}_5=49$
$\mathrm{ar}^2 \cdot \mathrm{ar}^4=49$
$\mathrm{a}^2 \mathrm{r}^6=49$
$\mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3}$
$\mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3}$
$\frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t}$
$\frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3}$
$3 \mathrm{t}^2-10 \mathrm{t}+3=0$
$\mathrm{t}=3, \frac{1}{3}$
Increasing G.P.
$\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$
$\mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8$
$=\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7$
$=\operatorname{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right)$
$=7(1+3+9)=91$
$\mathrm{ar}+\mathrm{ar}^5=\frac{70}{3}$
$\mathrm{~T}_3 \cdot \mathrm{T}_5=49$
$\mathrm{ar}^2 \cdot \mathrm{ar}^4=49$
$\mathrm{a}^2 \mathrm{r}^6=49$
$\mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3}$
$\mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3}$
$\frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t}$
$\frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3}$
$3 \mathrm{t}^2-10 \mathrm{t}+3=0$
$\mathrm{t}=3, \frac{1}{3}$
Increasing G.P.
$\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$
$\mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8$
$=\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7$
$=\operatorname{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right)$
$=7(1+3+9)=91$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.