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# Find out whether the series 13, 19, 25, 31, ... form an A.P.

Last updated date: 10th Sep 2024
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Hint: We need to form the general terms. The terms in the series should have a common difference between two consecutive terms. We find both to find out if the given series satisfies the conditions of an A.P.

Complete step-by-step solution:
The arithmetic progression is a part of a series. It consists mainly of an initial number or the starting number from where the progression begins. We call that term as “a”. Then we need a fixed number called the common difference which is added every time with a number to get the next term of the series. We take the fixed number as d.
We name every term of the progression as ${{t}_{n}}$ where n is the position of the number in the series.
So, ${{t}_{1}}$ is the first number i.e. the initial number which means ${{t}_{1}}=a$.
We add d to the first number to get the next one i.e. ${{t}_{2}}$. So, ${{t}_{2}}=a+d$.
We keep on getting the numbers of the series just by adding the common difference d.
So, ${{t}_{3}}=a+d+d=a+2d$, ${{t}_{4}}=a+2d+d=a+3d$, ${{t}_{5}}=a+3d+d=a+4d$ …..
We get all the terms of the series. Now, we find the general term.
The term will ${{t}_{n}}=a+(n-1)d$.
So, to find if a given series satisfies the rules of an A.P. we need to find out the common difference of the series and see if we can form a general form of the series.
Here the series is 13, 19, 25, 31, ...
We name the terms as ${{t}_{1}}=13$, ${{t}_{2}}=19$, ${{t}_{3}}=25$, ${{t}_{4}}=31$, …..
We find the differences between two consecutive terms
${{t}_{2}}-{{t}_{1}}=19-13=6$
${{t}_{3}}-{{t}_{2}}=25-19=6$
${{t}_{4}}-{{t}_{3}}=31-25=6$
We see the common difference = d is 6.
Now we find the general term of the series.
We know ${{t}_{n}}=a+(n-1)d$.
The initial term is 13 as the series is starting from 13 which means $a=13$.
So, ${{t}_{n}}=a+(n-1)d=13+6(n-1)=6n+7$.
We get the terms by putting values of n in the general term form.
${{t}_{1}}=6\times 1+7=13$
${{t}_{2}}=6\times 2+7=19$
${{t}_{3}}=6\times 3+7=25$
${{t}_{4}}=6\times 4+7=31$
So, the series 13, 19, 25, 31, ... forms an A.P.

Note: The common difference is more important than the general terms. The general term form is created with the initial term and the common difference. So, the common difference mainly decides whether the series is an A.P or not.