Chapter 12.2, Problem 25E

To determine

Calculate the first five terms of the given sequence $a_n=6n-10$ 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 25E

The first five term of the given sequence $a_n=6n-10$ is $-4,2,8,14\&20$

The given sequence in arithmetic .

Having the common difference, $d=6$

The $n^{th}$ term of the sequence in the standard form $a_n=a+(n-1)d$ is $a_n=-4+(n-1)6$

Explanation of Solution

Given information:

It is given in the question that,

the given sequence is $a_n=6n-10$

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 $a_n=6n-10$

So,

By putting $n=1$ in $a_n=6n-10$

  $\begin{array}{l}{a}_1=6*1-10\\ =6-10\\ =4\end{array}$

By putting $n=2$ in $a_n=6n-10$

  $\begin{array}{l}{a}_2=6*2-10\\ =12-10\\ =2\\ \end{array}$

By putting $n=3$ in $a_n=6n-10$

  $\begin{array}{l}{a}_3=6*3-10\\ =18-10\\ =8\end{array}$

By putting $n=4$ in $a_n=6n-10$

  $\begin{array}{l}{a}_4=6*4-10\\ =24-10\\ =14\end{array}$

By putting $n=5$ in $a_n=6n-10$

  $\begin{array}{l}{a}_5=6*5-10\\ =30-10\\ =20\end{array}$

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

  $\begin{array}{l}d\Rightarrow a_2-a_1=a_3-a_2\\ \Rightarrow 2-(-4)=8-2\\ \Rightarrow 2+4=6\\ \Rightarrow 6=6\end{array}$

Here, the sequence satisfied the given formula concept so the given sequence $a_n=6n-10$ which shows that the common difference of two consecutive term is same.

So, the given sequence in arithmetic.

Having the common difference, $d=6$

The $n^{th}$ term of the sequence in the standard form $a_n=a+(n-1)d$ is $a_n=-4+(n-1)6$