Question

Two AP’s have the same common difference. The difference between their ${{100}^{th}}$ terms is 100, what is the difference between their ${{1000}^{th}}$ terms?

Answer 1

We start solving the problem by assuming the first term and common difference for both AP’s. We then recall the definition ${{n}^{th}}$ term of AP and the formula as ${{T}_{n}}=a+\left( n-1 \right)d$ to find the ${{100}^{th}}$ terms in both AP’s. We then take the difference of these ${{100}^{th}}$ terms and equate it to 100 to get our first relation. We then find the ${{1000}^{th}}$ terms in both AP’s and take the difference of them to get the required result by using the relation we obtained.

Complete step-by-step answer:
According to the problem, we are given that two AP’s have the same common difference. We need to find the difference between their ${{1000}^{th}}$ terms if the difference between their ${{100}^{th}}$ terms is 100.
Let us assume the first term and common difference of the first AP be ${{a}_{1}}$ and ${{d}_{1}}$, and the first term and common difference of the second AP be ${{a}_{2}}$ and ${{d}_{1}}$.
We know that general equation of the ${{n}^{th}}$ term in an AP with first term ‘a’ and common difference ‘d’ is ${{T}_{n}}=a+\left( n-1 \right)d$. Let us find the ${{100}^{th}}$ term in both AP’s using this.
So, we get ${{100}^{th}}$ term of first AP as ${{T}_{{{100}_{1}}}}={{a}_{1}}+\left( 100-1 \right){{d}_{1}}={{a}_{1}}+99{{d}_{1}}$.
Similarly, we get ${{100}^{th}}$ term of first AP as ${{T}_{{{100}_{2}}}}={{a}_{2}}+\left( 100-1 \right){{d}_{1}}={{a}_{2}}+99{{d}_{1}}$.
According to the problem, we are given that the difference between ${{100}^{th}}$ terms of both AP’s is 100.
So, we have ${{T}_{{{100}_{1}}}}-{{T}_{{{100}_{2}}}}=100$.
$\Rightarrow {{a}_{1}}+99{{d}_{1}}-\left( {{a}_{2}}+99{{d}_{1}} \right)=100$.
$\Rightarrow {{a}_{1}}+99{{d}_{1}}-{{a}_{2}}-99{{d}_{1}}=100$.
$\Rightarrow {{a}_{1}}-{{a}_{2}}=100$ ---(1).
Now, let us find the ${{1000}^{th}}$ term in both AP’s.
So, we get ${{1000}^{th}}$ term of first AP as ${{T}_{{{1000}_{1}}}}={{a}_{1}}+\left( 1000-1 \right){{d}_{1}}={{a}_{1}}+999{{d}_{1}}$.
Similarly, we get ${{1000}^{th}}$ term of first AP as ${{T}_{{{1000}_{2}}}}={{a}_{2}}+\left( 1000-1 \right){{d}_{1}}={{a}_{2}}+999{{d}_{1}}$.
Now, let us find the between ${{1000}^{th}}$ terms of both AP’s.
So, we have ${{T}_{{{1000}_{1}}}}-{{T}_{{{1000}_{2}}}}={{a}_{1}}+999{{d}_{1}}-\left( {{a}_{2}}+999{{d}_{1}} \right)$.
$\Rightarrow {{T}_{{{1000}_{1}}}}-{{T}_{{{1000}_{2}}}}={{a}_{1}}+999{{d}_{1}}-{{a}_{2}}-999{{d}_{1}}$.
$\Rightarrow {{T}_{{{1000}_{1}}}}-{{T}_{{{1000}_{2}}}}={{a}_{1}}-{{a}_{2}}$.
From equation (1) we get ${{T}_{{{1000}_{1}}}}-{{T}_{{{1000}_{2}}}}=100$.
So, we have found the between the ${{1000}^{th}}$ terms of both AP’s as 100.

Note: We should not just randomly say that the difference between the ${{1000}^{th}}$ terms is 1000 without calculations. We need to use each and every detail of information given in the problem to get the required answer. We can see that the difference ${{r}^{th}}$ terms of both AP’s is equal to the difference of first terms of both AP’s which is constant all the time. Similarly, we can expect problems involving geometric progressions.