Question

In the list of 4, 9, 14, 19, 24, …. the first term is 4 and each term thereafter is 5 more than the previous term. Calculate the difference between the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term.
A. 5
B. 15
C. 20
D. 25
E. 30

Answer 1

Hint: From the given series of A.P. we find the general term of the series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. We put the values of n as 5790 and 5795 to find the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term. Then we find the difference between the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term.

Complete step-by-step answer:
We have been given a series of A.P. which is 4, 9, 14, 19, 24, ….
We express the A.P. in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term is 4. So, ${{t}_{1}}=4$. The common difference is 5. So, $d=5$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=4+5\left( n-1 \right)=5n-1$.
Now we need to find the difference between the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term.
We put two values of n as 5790 and 5795.
The ${{5790}^{th}}$ term is ${{t}_{5790}}=5\times 5790-1$.
The ${{5795}^{th}}$ term is ${{t}_{5795}}=5\times 5795-1$.
The difference is ${{t}_{5795}}-{{t}_{5790}}$. Putting the values, we get
${{t}_{5795}}-{{t}_{5790}}=\left( 5\times 5795-1 \right)-\left( 5\times 5790-1 \right)=5\left( 5795-5790 \right)=25$.
Therefore, the difference between the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term is 25.

So, the correct answer is “Option D”.

Note: We also can find the solution using the difference or gap between two numbers. As we know the common difference is 5, we try to find the gap between the ${{5790}^{th}}$ term and the ${{5795}^{th}}$ term. Then we try to take the number of gaps which is 5 and multiply that with the common difference to find the solution of the problem.