Question

Find the $5th$ term of the G.P. $\dfrac{5}{2},1,....$

Answer 1

Hint: First, before proceeding for this, we must know the formula for the $nth$ term of G.P. which is given by ${{a}_{n}}=a{{r}^{n-1}}$. Then, to get the $5th$ of the given series, we must find the value of r and a. Then, by subsisting the value of n as 5, a as $\dfrac{5}{2}$and r as $\dfrac{2}{5}$, we get the value of $5th$term of G.P.

Complete step-by-step answer:
In this question, we are supposed to find the $5th$ term of the G.P. $\dfrac{5}{2},1,....$
So, before proceeding for this, we must know the formula for the $nth$ term of G.P. which is given by:
${{a}_{n}}=a{{r}^{n-1}}$
Here, above a is the first term of geometric progression and r is the common ratio of the terms which is mostly given by $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}$ where ${{a}_{2}}$ is second term and ${{a}_{1}}$ is the first term of the given series.
So, to get the $5th$ of the given series, we must find the value of r and a.
Now, we can see clearly that the value of a is $\dfrac{5}{2}$ and value of r is given by:
$\begin{align}
  & r=\dfrac{1}{\dfrac{5}{2}} \\
 & \Rightarrow r=1\times \dfrac{2}{5} \\
 & \Rightarrow r=\dfrac{2}{5} \\
\end{align}$
So, after getting the value of a as $\dfrac{5}{2}$ and r as $\dfrac{2}{5}$, we can get the value of $5th$ term very easily.
Now, by subsisting the value of n as 5, a as $\dfrac{5}{2}$ and r as $\dfrac{2}{5}$, we get the value of $5th$term of G.P as:
${{a}_{5}}=\dfrac{5}{2}\times {{\left( \dfrac{2}{5} \right)}^{5-1}}$
Now, by solving the above expression, we get the desired result as:
$\begin{align}
  & {{a}_{5}}=\dfrac{5}{2}\times {{\left( \dfrac{2}{5} \right)}^{4}} \\
 & \Rightarrow {{a}_{5}}=\dfrac{5}{2}\times \dfrac{16}{625} \\
 & \Rightarrow {{a}_{5}}=\dfrac{8}{125} \\
\end{align}$
So, we get the $5th$term of G.P $\dfrac{5}{2},1,....$ as $\dfrac{8}{125}$.
Hence, $\dfrac{8}{125}$is the correct answer.

Note: Now, to solve these type of the questions we need to be careful with the type of progression given or mentioned as sometimes we mix geometric progression with arithmetic progression and uses the formula for nth term as ${{a}_{n}}=a+\left( n-1 \right)d$ which is not correct for this question as G.P is asked.