Question

Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Answer 1

Hint: Here, we can form equations according to the conditions given in the question and then we will try to find a relation between both the AP’s. We will use this relation to find the difference between their 1000th terms. We will also use the expression for any general term of an AP given as ${{a}_{n}}=a+\left( n-1 \right)d$.

Complete step-by-step answer:
Since, it is given that the common differences of both the AP’s are the same. So, let us consider the common difference to be equal to d.
Let the first term of the first AP be = ${{a}_{1}}$
 Let the first term of the second AP be = ${{a}_{2}}$
We know that the in general nth term of an AP is given as:
${{a}_{n}}=a+\left( n-1 \right)d$

So, using this expression for the general term of an AP, the 100^th term of the first AP will be:
$ ={{a}_{1}}+\left( 100-1 \right)d $
$ ={{a}_{1}}+99d $
And again using this general form, the 100th term of the second AP will be:
$ ={{a}_{2}}+\left( 100-1 \right)d $
$ ={{a}_{2}}+99d $
Since, it is given that the difference between the 100^th terms of these two AP’s is 100. So, we can write the following equation:
$ {{a}_{1}}+99d-\left( {{a}_{2}}+99d \right)=100 $
$ {{a}_{1}}+99d-{{a}_{2}}-99d=100 $
$ {{a}_{1}}-{{a}_{2}}=100 $..........................(1) 

Now, 1000^th term of the first AP is:
$ ={{a}_{1}}+\left( 1000-1 \right)d $
$ ={{a}_{1}}+999d $
And, 1000^th term of the second AP is:
$ ={{a}_{2}}+\left( 1000-1 \right)d $
$ ={{a}_{2}}+999d $
So, the difference between their 1000^th terms can be given as:
$ ={{a}_{1}}+999d-\left( {{a}_{2}}+999d \right) $
$ ={{a}_{1}}+999d-{{a}_{2}}-999d $
$  ={{a}_{1}}-{{a}_{2}} $

On substituting the value from equation (1), we have:
${{a}_{1}}-{{a}_{2}}=100$
Hence, the difference between the 1000^th terms of the given AP’s is 100.

Note: Students should note here that as the common difference of both the AP’s are same, so the difference between any of their nth terms will always be equal to the difference between their first terms.