Question

In a GP, the $3\mathrm{rd}$ term Is $24$ and the $6\mathrm{th}$ term Is $192$ then find the $10\mathrm{th}$ term.

Answer 1

Step 1: Form equation using given terms of Geometric Progression

The $3\mathrm{rd}$ term $\left(a_3\right)=24$

And, the $6\mathrm{th}$ term $\left(a_6\right)=192$

As we know, the $n^{th}$ term of a GP is defines as,

$a_n=a{r}^{n-1}$

So, the $3\mathrm{rd}$ term is,

$a_3=a{r}^2$

So, $a{r}^2=24$ $...\left(\mathrm{i}\right)$

And, the $6\mathrm{th}$ term is,

$a_6=a{r}^5$

So, $a{r}^5=192$ $...\left(\mathrm{ii}\right)$

Step 2: Calculate the common ratio $r$

On dividing the equation $\left(\mathrm{ii}\right)$ by equation $\left(i\right)$, we get,

$\frac{a{r}^5}{a{r}^{2^{}}}=\frac{192}{24}$

$\Rightarrow$ $\frac{r^5}{r^2}=\frac{8}{1}$

$\Rightarrow$ $r^{5-2}=8$ $\left[\because \frac{a^m}{a^n}=a^{m-n}\right]$

$\Rightarrow$ $r^3=8$

$\Rightarrow$ $\sqrt[3]{r^3}=\sqrt[3]{8}$ [Taking cube root on both sides]

$\Rightarrow$ $\sqrt[3]{r^3}=\sqrt[3]{2^3}$

$\Rightarrow$ $r=2$ $[\because \sqrt[n]{{\left(a\right)}^n}=a]$

Step 3: Calculate the first term $a$

Substituting the value of $r$ in equation $\left(\mathrm{ii}\right)$,

$\because$ $a{r}^2=24$

$\Rightarrow$ $a{\left(2\right)}^2=24$

$\Rightarrow$ $4a=24$

$\Rightarrow$ $a=\frac{24}{4}$

$\Rightarrow$ $a=6$

Step 4: Calculate the $10\mathrm{th}$ term

As we know, the $n^{th}$ term of a GP is defines as,

$a_n=a{r}^{n-1}$

So, the $10\mathrm{th}$ term is,

$a_{10}=a{r}^9$

$\Rightarrow$ $a_{10}=\left(6\right){\left(2\right)}^9$ $\left[\because a=3,r=2\right]$

$\Rightarrow$ $a_{10}=\left(3\right)\left(2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\right)$

$\Rightarrow$ $a_{10}=3\times 1024$

$\Rightarrow$ $a_{10}=3072$

Hence, the $10\mathrm{th}$ term of the given GP is $3072$.