Question

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

Answer 1

• Step 1: Finding out the difference between their ${1000}^{th}$ terms.

Let, the first term of two $A{P}_S$s be $a_1$ and $a_2$ respectively

And the common difference between these $A{P}_S$ be $d$.

For the first A.P.,we know,

$a_n=a+(n-1)d$

Therefore,

$$\begin{gathered} a_{100}=a_1+(100-1)d \\ =a_1+99d \\ {a}_{1000}=a_1+(1000-1)d \\ =a_1+\left(999d\right) \end{gathered}$$

• Step 2: Finding out the difference between their ${1000}^{th}$ terms.

For second A.P., we know,

$a_n=a+(n-1)d$

Therefore,

$$\begin{gathered} a_{100}=a_2+(100-1)d \\ {a}_2+99d \\ {a}_{1000}=a_2+(1000-1)d \\ {a}_2+999d \end{gathered}$$

Given that, difference between ${100}^{th}$ term of the two $APs=100$

Therefore, $$\begin{gathered} (a_1+99d)-(a_2+99d)=100 \\ {a}_1-a_2=100\to \left(1\right) \end{gathered}$$

Difference between ${1000}^{th}$ terms of the two $APs$

$$\begin{gathered} \left(a_1+999d\right)-(a_2+999d)=a_1-a_2 \\ Fromequation\left(1\right) \end{gathered}$$

This difference, $a_1-a_2=100$

Hence, the difference between ${1000}^{th}$ terms of the two $A.P.$ will be $100.$