Chapter 3.6, Problem 40PFA

To determine

To Find: The value of $n^{th}$ term of an arithmetic progression.

Answer to Problem 40PFA

  $n^{th}$ term of an arithmetic progression is $a_n=12-5n$

Explanation of Solution

Given information:

First four terms of an arithmetic sequence is $17,22,27\&32$

Concept and Formula Used:

Arithmetic Sequence: The sequence with difference between successive terms is constant.

  $n^{th}$ term of an arithmetic progression is $a_n=a+\left(n-1\right)d$

Where ‘a’ is the first term and ‘d’ is the common difference.

Calculation:

It is given that four terms of an arithmetic sequence are $17,22,27\&32$

As in arithmetic sequence the difference between successive terms is constant.

Therefore $d=22-17=5\&a=17$

The $n^{th}$ term of an arithmetic progression is

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ a_n=17+\left(n-1\right)\left(5\right)\\ a_n=17+5n-5\\ a_n=12-5n\end{array}$

Conclusion:

  $n^{th}$ term of an arithmetic progression is $a_n=12-5n$