# The sum of first four terms of a GP is 30 and that of last four terms is 960. If the first and the last term of the GP is 2 and 512 respectively, find the common ratio.?

Jul 28, 2018

$2 \sqrt[3]{2}$.

#### Explanation:

Suppose that the common ratio (cr) of the GP in question is $r$ and ${n}^{t h}$

term is the last term.

Given that, the first term of the GP is $2$.

$\therefore \text{The GP is } \left\{2 , 2 r , 2 {r}^{2} , 2 {r}^{3} , . . , 2 {r}^{n - 4} , 2 {r}^{n - 3} , 2 {r}^{n - 2} , 2 {r}^{n - 1}\right\}$.

Given, $2 + 2 r + 2 {r}^{2} + 2 {r}^{3} = 30. . . \left({\star}^{1}\right) , \mathmr{and} ,$

$2 {r}^{n - 4} + 2 {r}^{n - 3} + 2 {r}^{n - 2} + 2 {r}^{n - 1} = 960. . . \left({\star}^{2}\right)$.

We also know that the last term is $512$.

$\therefore {r}^{n - 1} = 512. \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}^{3}\right)$.

Now, $\left({\star}^{2}\right) \Rightarrow {r}^{n - 4} \left(2 + 2 r + 2 {r}^{2} + 2 {r}^{3}\right) = 960 ,$

$i . e . , \frac{{r}^{n - 1}}{r} ^ 3 \left(2 + 2 r + 2 {r}^{2} + 2 {r}^{3}\right) = 960$.

:. (512)/r^3(30)=960......[because, (star^1) & (star^3)].

$\therefore r = \sqrt[3]{512 \cdot \frac{30}{960}} = 2 \sqrt[3]{2}$, is the desired (real) cr!