# In an increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th terms is equal to :   Option: 1 32 Option: 2 26 Option: 3 30 Option: 4 35

$\\\text {Let a, ar, ar }^{2}, \ldots\text{are in GP}\\\text{according to question}\\ \mathrm{T}_{2}+\mathrm{T}_{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^{4}\right)=\frac{25}{2}$

\begin{aligned} &a^{2} r^{2}\left(1+r^{4}\right)^{2}=\frac{625}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)\\ &\mathrm{T}_{3} \cdot \mathrm{T}_{5}=25 \Rightarrow\left(\mathrm{ar}^{2}\right)\left(\mathrm{ar}^{4}\right)=25\\ &a^{2} r^{6}=25\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2) \end{aligned}

on dividing (1) by (2), we get

$\\\frac{\left(1+r^{4}\right)^{2}}{r^{4}}=\frac{25}{4}\\ 4 r^{8}-17 r^{4}+4=0 \\ \left(4 r^{4}-1\right)\left(r^{4}-4\right)=0 \\ r^{4}=\frac{1}{4}, 4 \Rightarrow r^{4}=4$

For an increasing GP

\\ \mathrm{a}^{2} \mathrm{r}^{6}=25 \\\Rightarrow\left(\mathrm{ar}^{3}\right)^{2}=25 \\ \begin{aligned} \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) \\ &=5(1+2+4)=35 \end{aligned}

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