Chapter 6.5, Problem 66CCR

To determine

To find: Next two terms in the sequence.

Answer to Problem 66CCR

Next two terms of the series are $62.5,156.25$

Explanation of Solution

Given information:

  $1.6,4,10,25,\mathrm{....}$

Concept:

A Geometric sequence multiplies a fixed amount (common ratio) to get the next term in the sequence.

We know $n^{th}$ term of Geometric progression is

  $a_n=a{r}^{n-1}$

Where

  $a$ = First term

  $r$ = Common ratio.

Calculation:

Given series is

  $1.6,4,10,25,\mathrm{....}$

Common ratio= $r=\frac{4}{1.6}=\frac{10}{4}=\frac{25}{10}=2.5$

From given sequence

First term = $a=1.6$

Common ratio = $r=2.5$

Substituting values in $n^{th}$ term formula

We get

  $\begin{array}{l}{a}_n=a{r}^{n-1}\\ \Rightarrow a_n=\left(1.6\right){\left(2.5\right)}^{n-1}\end{array}$

Therefore,

  $n^{th}$ term of the sequence is $a_n=\left(1.6\right){\left(2.5\right)}^{n-1}$

Given four terms so we need to find fifth and sixth terms.

Substitute $n=5$

  $\begin{array}{l}{a}_n=\left(1.6\right){\left(2.5\right)}^{n-1}\\ \Rightarrow a_5=\left(1.6\right){\left(2.5\right)}^{5-1}\\ \Rightarrow a_5=\left(1.6\right){\left(2.5\right)}^4\\ \Rightarrow a_5=\left(1.6\right)\left(39.0625\right)\\ \Rightarrow a_5=62.5\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_n=\left(1.6\right){\left(2.5\right)}^{n-1}\\ \Rightarrow a_6=\left(1.6\right){\left(2.5\right)}^{6-1}\\ \Rightarrow a_6=\left(1.6\right){\left(2.5\right)}^5\\ \Rightarrow a_6=\left(1.6\right)\left(97.65625\right)\\ \Rightarrow a_6=156.25\end{array}$

Therefore,

Next two terms of the series are $62.5,156.25$