Chapter 9, Problem 49RE

To determine

The $n^{th}$ term of the geometric sequence whose first two terms are “ $a_1=18\text{and}{a}_2=-9$ ” and find ${10}^{th}$ term of the sequence.

Answer to Problem 49RE

The $n^{th}$ and ${10}^{th}$ terms of given geometric sequence are $18{\left(-\frac{1}{2}\right)}^{n-1}$ and $-\frac{9}{256}$ respectively.

Explanation of Solution

Given information :

Given first two terms “ $a_1=18\text{and}{a}_2=-9$ ” of geometric sequence.

As we know that common ratio $r=\frac{a_2}{a_1}$ ,

  $\begin{array}{l}\Rightarrow r=\frac{-9}{18}\\ \therefore r=\left(-\frac{1}{2}\right)\end{array}$

Thus, the $n^{th}$ term of geometric sequence is $a_n=a_1{\left(r\right)}^{n-1},\text{where}{a}_1=\text{first}\text{term}andr=\text{common}\text{ratio}$

  $\begin{array}{l}\Rightarrow a_n=18{\left(-\frac{1}{2}\right)}^{n-1}\\ forn=10,\\ \Rightarrow a_{10}=18{\left(-\frac{1}{2}\right)}^{10-1}\\ =18{\left(-\frac{1}{2}\right)}^9\\ \therefore a_{10}=-\frac{9}{256}\end{array}$

Hence, the $n^{th}$ and ${10}^{th}$ terms of given geometric sequence are $18{\left(-\frac{1}{2}\right)}^{n-1}$ and $-\frac{9}{256}$ respectively.