Question

 Find the \[{{n}^{th}}\] term of a GP \[\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},.......\]

Answer 1

Hint: Find the first term and common difference of the given GP. Then substitute these values in the formula to find the \[{{n}^{th}}\] term of the GP.

Complete Step-by-Step solution:
Geometric Progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio (r).
Now we have been given the geometric progression,
\[\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},.......\]
We know the first term, \[a=\dfrac{3}{2}\].
To find common ratio = \[{{2}^{nd}}\] term / \[{{1}^{st}}\] term = \[\dfrac{\dfrac{3}{4}}{\dfrac{3}{2}}=\dfrac{3}{4}\times \dfrac{2}{3}=\dfrac{1}{2}\].
Similarly, r = \[{{3}^{rd}}\] term / \[{{2}^{nd}}\] term = \[\dfrac{\dfrac{3}{8}}{\dfrac{3}{4}}=\dfrac{3}{8}\times \dfrac{4}{3}=\dfrac{1}{2}\].
Thus we got the first term = \[a=\dfrac{3}{2}\].
Common ratio \[=r=\dfrac{1}{2}\].
The \[{{n}^{th}}\] term of a GP series is given by, \[{{a}_{n}}=a{{r}^{n-1}}\].
We know, \[a=\dfrac{3}{2}\], \[r=\dfrac{1}{2}\].
\[\therefore {{n}^{th}}\] term of the given GP = \[{{a}_{n}}=a{{r}^{n-1}}\]\[=\left( \dfrac{3}{2} \right){{\left( \dfrac{1}{2} \right)}^{n-1}}\].
\[\begin{align}
  & {{a}_{n}}=\dfrac{3}{2}\times \dfrac{{{\left( \dfrac{1}{2} \right)}^{n-1}}}{\dfrac{1}{2}}=\dfrac{3}{2}\times \dfrac{2}{1}\times {{\left( \dfrac{1}{2} \right)}^{n-1}} \\
 & {{a}_{n}}=3{{\left( \dfrac{1}{2} \right)}^{n-1}} \\
\end{align}\]
Thus we got the \[{{n}^{th}}\] term of the GP, \[{{a}_{n}}=3{{\left( \dfrac{1}{2} \right)}^{n-1}}\].

Note: If we were asked to find the value of \[{{4}^{th}}\] term, then
\[{{a}_{n}}=3{{\left( \dfrac{1}{2} \right)}^{n-1}}\Rightarrow {{a}_{4}}=3{{\left( \dfrac{1}{2} \right)}^{4}}=3\left[ \dfrac{1}{{{2}^{4}}} \right]=\dfrac{3}{16}\]
Similarly, \[{{5}^{th}}\] term \[{{a}_{5}}=3{{\left( \dfrac{1}{2} \right)}^{5}}=\dfrac{3}{32}\] and the series can go on like this.