Question

The 4th term of a geometric progression is $\frac{2}{3}$ and 7th term is $\frac{16}{81}$. Find the Geometric series.

• A. $\frac{9}{4},-\frac{3}{2},1,-\frac{2}{3},......$
• B. $\frac{9}{4},\frac{3}{2},1,\frac{2}{3},......$
• C. $-\frac{9}{4},\frac{3}{2},1,\frac{2}{3},......$
• D. None of these

Answer 1

The correct option is B $\frac{9}{4},\frac{3}{2},1,\frac{2}{3},......$

Given: $a_4$ of GP $=\frac{2}{3},a_7=\frac{16}{81}$

We know that $n^{th}$ term of a GP is given by $a_n=a{r}^{n-1}$, where $a=$first term, $r=$ common ratio

$\therefore a_4=\frac{2}{3}$

$a{r}^{4-1}=\frac{2}{3}$

$\Rightarrow a{r}^3=\frac{2}{3}\dots \dots \left(i\right)$

And, $a_7=a{r}^{7-1}=\frac{16}{81}$

$\Rightarrow a{r}^6=\frac{16}{81}\dots \dots \left(ii\right)$

on dividing eq (i) by (ii), we get

$\frac{a{r}^3}{a{r}^6}=\frac{2\times 81}{3\times 16}$

$\Rightarrow \frac{1}{r^3}=\frac{27}{8}={\left(\frac{3}{2}\right)}^3$

$\Rightarrow r=\frac{2}{3}$

On putting the value of ${}^{\prime }{r}^{\prime }$ in eq.(i), we get

$a{\left(\frac{2}{3}\right)}^3=\left(\frac{2}{3}\right)$

$\Rightarrow a={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

$G.P=a,ar,a{r}^2,a{r}^3,.......$

$=\frac{9}{4},\frac{3}{2},1,\frac{2}{3},......$