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  • In an increasing geometric series, the sum of the second and the sixth term is and the product of the third and fifth term is 25. Then, the sum of 4th

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

Answers (1)

best_answer

\\\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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himanshu.meshram

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