Question

Question 12
Two APs 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

Let the first term of these A.Ps be m and n respectively and the common difference of these A.Ps be d.

For the first A.P,
$a_{100}=m+(100-1)d$
$a_{100}=m+99d...\left(i\right)$
$a_{1000}=m+(1000-1)d$
$a_{1000}=m+999d...\left(ii\right)$

For the second A.P,
$b_{100}=n+(100-1)d$
$b_{100}=n+99d...\left(iii\right)$
$b_{1000}=n+999d...\left(iv\right)$

Given that, the difference between the ${100}^{th}$ terms is 100
$\Rightarrow a_{100}-b_{100}=m+99d-n-99d$
$\Rightarrow 100=m-n...\left(v\right)$

Hence, the difference between the ${1000}^{th}$ terms is
$a_{1000}-b_{1000}=m+999d-n-999d$
$=m-n(fromeq.(v\left)\right)$
$=100$