Question

Two $AP\text{'}s$ have the same common difference. The first term of one $AP$ is $2$ and that of the other is $7$. The difference between their ${10}^{th}$ terms is the same as the difference between their ${21}^{st}$ terms, which is the same as the difference between any two corresponding terms. Why?

Answer 1

Let's find the difference between any two corresponding terms in an $AP$

Let us there are two $AP\text{'}s$ with first terms $a$ and $A$

Let their common differences be $d$ and $D$ respectively

Suppose $n$ be any term

$$\begin{gathered} a_n=a+(n-1)d \\ {A}_n=A+(n-1)D \end{gathered}$$

As the common difference is equal for both $AP\text{'}s$

We have $D=d$

Using this we can solve the equation.

We know that the $n^{th}$ term of an $AP$ is given by the formula

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

where,

$a=$ first term

$a_n$ is the nth term

$d$ is the common difference

So,

$$\begin{gathered} A_n-a_n=a+(n-1)d-\left[A+\left(n-1\right)D\right] \\ =a+\left(n-1\right)d-A-(n-1)d \\ =a-A \end{gathered}$$

As $a-A$ is a constant value

Therefore, the difference between any corresponding terms will be equal to $a-A$.