If the sum and product of four positive consecutive terms of a G.P.,n are 126 and 1296, respectively, then the sum of common ratios of alln such GPs is

n a, ar, ar^2, ar^3( a, r>0) n a^4 r^6=1296 n a^2 r^3=36 n a=frac6 r^3 / 2 n a+ ar+ ar^2+ ar^3=126 n frac1 r^3 / 2+frac r r^3 / 2+frac r^2 r^3 / 2+frac r^3 r^3 / 2=(126 / 6)=21 n ( r^-3 / 2+ r^3 / 2)+( r^1 / 2+ r^-1 / 2)=21 n r^1 / 2+ r^-1 / 2= A n r^-3 / 2+ r^3 / 2+3 A= A^3 n A^3-3 A+ A=21 n A^3-2 A=21 n A=3 n sqrt r+frac1sqrt r=3 n r+1=3 sqrt r n r^2+2 r+1=9 r n r^2-7 r+1=0n\

If the sum and product of four positive consecutive ter... | Sequence & Series (Mathematics)

\begin{aligned}\n& \mathrm{a}, \operatorname{ar}, \mathrm{ar}^2, \mathrm{ar}^3(\mathrm{a}, \mathrm{r}>0) \\\n& \mathrm{a}^4 \mathrm{r}^6=1296 \\\n& \mathrm{a}^2 \mathrm{r}^3=36 \\\n& \mathrm{a}=\frac{6}{\mathrm{r}^{3 / 2}} \\\n& \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\mathrm{ar}^3=126 \\\n& \frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^2}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^3}{\mathrm{r}^{3 / 2}}=\frac{126}{6}=21 \\\n& \left(\mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}\right)+\left(\mathrm{r}^{1 / 2}+\mathrm{r}^{-1 / 2}\right)=21 \\\n& \mathrm{r}^{1 / 2}+\mathrm{r}^{-1 / 2}=\mathrm{A} \\\n& \mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}+3 \mathrm{~A}=\mathrm{A}^3 \\\n& \mathrm{~A}^3-3 \mathrm{~A}+\mathrm{A}=21 \\\n& \mathrm{~A}^3-2 \mathrm{~A}=21 \\\n& \mathrm{~A}=3 \\\n& \sqrt{\mathrm{r}}+\frac{1}{\sqrt{\mathrm{r}}}=3 \\\n& \mathrm{r}+1=3 \sqrt{\mathrm{r}} \\\n& \mathrm{r}^2+2 \mathrm{r}+1=9 \mathrm{r} \\\n& \mathrm{r}^2-7 \mathrm{r}+1=0\n\end{aligned}

Explanation

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