Find the GP whose 4th and 7th terms are

Question:

Find the GP whose $4^{\text {th }}$ and $7^{\text {th }}$ terms are $\frac{1}{18}$ and $\frac{-1}{486}$ respectively.

 

Solution:

The $n^{\text {th }}$ term of a GP is $a_{n}=a r^{n-1}$

It's given in the question that $4^{\text {th }}$ term of the GP is $\frac{1}{18}$ and $7^{\text {th }}$ term of GP is $-\frac{1}{486}$.

So, $a_{4}=a r^{3}=\frac{1}{18} \rightarrow(1)$

$a^{7}=a r^{6}=-\frac{1}{486} \longrightarrow(2)$

$\frac{(2)}{(1)} \rightarrow \frac{\operatorname{ar}^{6}}{\operatorname{ar}^{3}}=r^{3}=-\frac{1}{27}$

Common ratio, $\mathrm{r}=-\frac{1}{3}$

$a r^{3}=\frac{1}{18}$

$a=-\frac{3}{2}$

The required GP is of form $a, a r, a r^{2}, a r^{3}, a r^{4} \ldots .$

The first term of GP, $a=-\frac{3}{2}$

The second term of GP, ar $=-\frac{3}{2} \mathrm{x}-\frac{1}{3}=\frac{1}{2}$

The third term of GP, $\mathrm{ar}^{2}=\frac{1}{2} \mathrm{x}-\frac{1}{3}=-\frac{1}{6}$

The fourth term of GP, $\operatorname{ar}^{3}=-\frac{1}{6} \mathrm{x}-\frac{1}{3}=\frac{1}{18}$

The fifth term of $\mathrm{GP}, \mathrm{ar}^{4}=\frac{1}{18} \mathrm{x}-\frac{1}{3}=-\frac{1}{54}$

And so on...

The required GP is $-\frac{3}{2}, \frac{1}{2},-\frac{1}{6}, \frac{1}{18},-\frac{1}{54} \ldots \ldots . .$

 

 

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