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

n T_2+ T_6=(70 / 3) n ar+ ar^5=(70 / 3) n T_3 * T_5=49 n ar^2 * ar^4=49 n a^2 r^6=49 n ar^3=+7, a=frac7 r^3 n ar(1+ r^4)=(70 / 3) n frac7 r^2(1+ r^4)=(70 / 3), r^2= t n frac1 t(1+ t^2)=(10 / 3) n 3 t^2-10 t+3=0 n t=3, (1 / 3) n Increasing G.P. r^2=3, r=sqrt(3) n T_4+ T_6+ T_8 n = ar^3+ ar^5+ ar^7 n = ar^3(1+ r^2+ r^4) n =7(1+3+9)=91nn n

In an increasing geometric progression of positive term... | Sequence & Series (Mathematics)

\begin{aligned}\n& \mathrm{T}_2+\mathrm{T}_6=\frac{70}{3} \\\n& \mathrm{ar}+\mathrm{ar}^5=\frac{70}{3} \\\n& \mathrm{~T}_3 \cdot \mathrm{~T}_5=49 \\\n& \mathrm{ar}^2 \cdot \mathrm{ar}^4=49 \\\n& \mathrm{a}^2 \mathrm{r}^6=49 \\\n& \mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3} \\\n& \mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3} \\\n& \frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t} \\\n& \frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3} \\\n& 3 \mathrm{t}^2-10 \mathrm{t}+3=0 \\\n& \mathrm{t}=3, \frac{1}{3} \\\n& \text { Increasing G.P. } \mathrm{r}^2=3, \mathrm{r}=\sqrt{3} \\\n& \mathrm{~T}_4+\mathrm{T}_6+\mathrm{T}_8 \\\n& =\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7 \\\n& =\mathrm{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right) \\\n& =7(1+3+9)=91\n\n\end{aligned}\n

Explanation

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