The sum of the first four terms of a geometri | vedclass

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(D) Let the first four terms be $a, ar, ar^2, ar^3$.The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \qua

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$3$

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(D) Let the first four terms be $a, ar, ar^2, ar^3$.The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \quad (1)$.The sum of their reciprocals is $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{r^3+r^2+r+1}{ar^3} = \frac{1}{a} \frac{r^4-1}{r^3(r-1)} = \frac{65}{18} \quad (2)$.Dividing $(1)$ by $(2)$,we get $\frac{a\frac{r^4-1}{r-1}}{\frac{1}{a}\frac{r^4-1}{r^3(r-1)}} = a^2 r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$.Given the product of the first three terms is $a \cdot ar \cdot ar^2 = a^3 r^3 = 1$,which implies $ar = 1$,so $a = \frac{1}{r}$.Substituting $a = \frac{1}{r}$ into $a^2 r^3 = \frac{3}{2}$,we get $(\frac{1}{r})^2 r^3 = r = \frac{3}{2}$.Thus,$a = \frac{1}{3/2} = \frac{2}{3}$.The third term $\alpha = ar^2 = \frac{2}{3} 	imes (\frac{3}{2})^2 = \frac{2}{3} 	imes \frac{9}{4} = \frac{3}{2}$.Therefore,$2\alpha = 2 	imes \frac{3}{2} = 3$.

The sum of the first four terms of a geometric progression $(G.P.)$ is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of the first three terms of the $G.P.$ is $1$,and the third term is $\alpha$,then $2\alpha$ is ....... .

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