Chapter 7.9, Problem 3GP

To determine

To find the n^th term and 10^th term of the given geometric sequence.

Answer to Problem 3GP

  $\begin{array}{l}{a}_n=96{\left(\frac{1}{2}\right)}^{n-1}\\ a_{10}=0.1875\end{array}$

Explanation of Solution

Given:

  $96,48,24,12,\mathrm{.......}$

Concept used:

The n^th term of a geometric sequence with first term ‘a’ and common ratio ‘r’ is

  $a_n=a{r}^{n-1}$

Calculation:

  $13,52,208,832,\mathrm{.......}$

Given sequence is a geometric sequence with

  $a=96,r=\frac{48}{96}=\frac{1}{2}$

Therefore n^th term is

  $\begin{array}{l}{a}_n=a{r}^{n-1}\\ \Rightarrow a_n=96{\left(\frac{1}{2}\right)}^{n-1}\end{array}$

And 10^th term is,

  $a_{10}=a{r}^9=96{\left(\frac{1}{2}\right)}^9=\frac{96}{512}=0.1875$