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Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Given:
Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100.
To do: We have to find the difference between their 1000th terms.
Solution:
Let $a, a+d, a+2d,......$ and $p, p+d, p+2d,.......$ be the two A.P.s.
Therefore,
$a_{100}=a+(100-1)d$
$=a+99d$
$p_{100}=p+(100-1)d$
$=p+99d$
According to the question,
$a+99d-(p+99d)=100$
$a-p=100$.....(i)
$a_{1000}=a+(1000-1)d$
$=a+999d$
$p_{1000}=p+(1000-1)d$
$=p+999d$
Therefore,
$a_{1000}-p_{1000}=a+999d-(p+999d)$
$=a+999d-p-999d$
$=a-p$
$=100$ (From (i))
The difference between their 1000th terms is $100$.
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