Chapter 12.2, Problem 22E

To determine

Calculate the first five terms of the given sequence $a_n=4+2^n$ and find whether it is arithmetic. If it is arithmetic then also calculate the common difference, and express the $n^{th}$ term of the sequence in the standard form $a_n=a+(n-1)d$ .

Answer to Problem 22E

The first five term of the given sequence $a_n=a+(n-1)d$ is $6,8,12,20,36$

The given sequence is not in arithmetic .

Explanation of Solution

Given information:

It is given in the question that,

the given sequence is $a_n=4+2^n$

Formula used:

  $a_n=a+(n-1)d$ & $d\Rightarrow a_n-a_{n-1}=a_{n+1}-a_n$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ disthecommondifference\end{array}$

Calculation:

For finding the first five term put $n=1,2,3,4,5$ in the given sequence is $a_n=4+2^n$

So,

By putting $n=1$ in $a_n=4+2^n$

  $\begin{array}{l}{a}_1=4+(2^1)\\ =4+2\\ =6\end{array}$

By putting $n=2$ in $a_n=4+2^n$

  $\begin{array}{l}{a}_2=4+(2^2)\\ =4+4\\ =8\end{array}$

By putting $n=3$ in $a_n=4+2^n$

  $\begin{array}{l}{a}_3=4+(2^3)\\ =4+8\\ =12\end{array}$

By putting $n=4$ in $a_n=4+2^n$

  $\begin{array}{l}{a}_4==4+(2^4)\\ =4+16\\ =20\end{array}$

By putting $n=5$ in $a_n=4+2^n$

  $\begin{array}{l}{a}_5=4+(2^5)\\ =4+32\\ =36\end{array}$

By using the formula $d\Rightarrow a_n-a_{n-1}=a_{n+1}-a_n$

  $\begin{array}{l}d\Rightarrow a_2-a_1=a_3-a_2\\ \Rightarrow 8-6=12-6\\ \Rightarrow 2=6\end{array}$

Here, the sequence not satisfied the given formula concept so the given sequence $a_n=4+2^n$ which shows that the common difference of two consecutive term is not same.

So, the given sequence is not in arithmetic.