# Ex.5.2 Q12 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

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

## Text Solution

**What is Known:?**

Two APs with the same common difference and difference between their \(100^\rm{th}\) term.

**What is Unknown?**

Difference between their \(1000^\rm{th}\) term

**Reasoning:**

\({a_n} = a + \left( {n - 1} \right)d\) is the general term of AP. Where \({a_n}\) is the \(n\rm{th}\) term, \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

Let the first term of these A.P.s be \({a_1}\) and \({b_1}\) respectively and the Common difference of these A.P’s be \(d\)

For first A.P.,

\[\begin{align}{a_{100}} &= {a_1} + (100 - 1)d\\ &= {a_1} + 99{\rm{d}}\\{a_{1000}} &= {a_1} + (1000 - 1)d\\{a_{1000}} &= {a_1} + 999d\end{align}\]

For second A.P.,

\[\begin{align}{b_{100}} &= {b_1} + (100 - 1)d\\& = {b_1} + 99d\\{b_{1000}}& = {b_1} + (1000 - 1)d\\ &= {{\rm{b}}_1} + 999d\end{align}\]

Given that, difference between

\(100^\rm{th}\) term of these A.P.s \(= 100\)

Thus, we have

\(\begin{align}&\left( {{a_1} + 99d} \right) - \left( {{b_1} + 99d} \right)= 100\\&{a_1} - {b_1}= 100\quad \dots {\rm{Equation}}\left( 1 \right)\end{align}\)

Difference between \(1000^\rm{th}\) terms of these A.P.s

\(\begin{align}&\left( {{a_1} + 999d} \right)\! -\! \left( {{b_1}\! +\! 999d} \right) \\&=\! {a_1}\! - \!{b_1}\quad\dots{\rm{Equation}}\left( 2 \right)\end{align}\)

From equation (1) & Equation (2),

This difference, \({a_1} - {b_1} = 100\)

Hence, the difference between \(1000^\rm{th}\)^{ }terms of these A.P. will be \(100.\)