Sum of an infinite G.P. is 2 and sum of their cubes is 24, then 5th term of the G.P. is

Sum of an infinite G.P. is 2 and sum of their cubes is 24, then 5th term of the G.P. is

1. A

3/16

2. B

3/8

3. C

– 3/8

4. D

– 3/16

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Solution:

$\begin{array}{l}\frac{a}{1-r}=2,\frac{{a}^{3}}{1-{r}^{3}}=24\\ \therefore \frac{8\left(1-r{\right)}^{3}}{1-{r}^{3}}=24⇒\frac{1-2r+{r}^{2}}{1+r+{r}^{2}}=3\\ ⇒2{r}^{2}+5r+2=0⇒r=-2,-1/2\end{array}$

As $|r|<1,r=-1/2,a=3$

Now, ${t}_{5}=a{r}^{4}=3/16$

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