Question

Two arithmetic progressions 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

Given two A.P have same common difference =$d$

let the first term of first A.P be $a_1$

and the first term of second A.P be $a_1^{\prime }$

hence 100th term of first A.P is given by

$a_{100}=a_1+(100-1)d$

$⟹a_{100}=a_1+99d...eq\left(1\right)$

and $100$th term of second A.P is given by

$a_{100}^{\prime }=a{,}_1+(100-1)d$

$⟹a_{100}^{\prime }=a_1^{\prime }+99d...eq\left(2\right)$

and given that the difference between their 100th term is 100

hence, $a_{100}-a_{100}^{\prime }=100$

$⟹(a_1+99d)-(a_1^{\prime }+99d)=100$

$⟹a_1-a_1^{\prime }=\mathrm{100...}\dots .eq\left(3\right)$

$1000$th term of first A.P is

$a_{1000}=a_1=(1000-1)d$

$⟹a_{1000}=a_1+999d...\dots .eq\left(4\right)$

and 1000th term of second A.P is

$a_{1000}^{\prime }=a_1^{\prime }=(1000-1)d$

$⟹a_{1000}^{\prime }=a_1^{\prime }+999d...\dots .eq\left(4\right)$

now difference of their $1000$th term is given by

$a_{1000}-a_{1000}^{\prime }=(a_1+999d)-(a_1^{\prime }+999d)$

$a_{1000}-a_{1000}^{\prime }=a_1-a_1^{\prime }$

put value of $a_1-a_1^{\prime }=100$ from eq(3) in above equation we get

$a_{1000}-a_{1000}^{\prime }=100$

hence the difference between their $1000$th term is $100.$