Which term of the G.P.:2,22,4, 128? - Vidyadip

Question

Which term of the G.P.:$2,2\sqrt{2},4,\ \dots\text{is}128?$

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Answer

$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$ $\text{t}_\text{n}=\text{ar}^{\text{n}-1}$ $\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$ $\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$ $\text{t}_\text{n}=\text{ar}^{\text{n}-1}$ $\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$ $\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$ $\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$ $\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$ $10=(\text{n}-1)$ $\text{n}=11$ $\therefore\text{term is 11}^\text{th}.$ $2,2\sqrt{2},4,\ \dots\text{is}128?$ $\text{a}=2,\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{2\sqrt{2}}{2}=\sqrt{2},\text{n}=?$ $\text{t}_\text{n}=128$ Also, $\text{t}_\text{n}=\text{ar}^{\text{n}-1}$ $128=(2)\big(\sqrt{2}\big)^{\text{n}-1}$ $\frac{128}{2}=\big(\sqrt{2}\big)^{\text{n}-1}$ $64=\big(\sqrt{2}\big)^{\text{n}-1}$ $(2)^6=\big(\sqrt{2}\big)^{\text{n}-1}$ $\Rightarrow12={\text{n}-1}$ ${\text{n}-13}$ $\therefore13^{\text{th}}\text{ term is 128}.$

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