Question
Two APs have the same common difference. The difference between their 100th terms is $100$, what is the difference between their $1000th$ terms?
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Answer
Let the common difference of two AP's be d, their first terms as a and a' nth term of both the AP's will be given by
$a_n=a+(n-1) d \text { and }$
$a_n^{\prime}=a^{\prime}+(n-1) d, \text { respectively }$
Now 100 th term of 1 st AP will be given by: $a_{100}=a+(100-1) d=a+99 d$
100th term of second AP will be given by: $a^{\prime}{ }_{100}=a+(100-1) d=a^{\prime}+99 d$
Now given, $a _{100}- a ^{\prime}{ }_{100}=( a +99 d)-\left( a ^{\prime}+99 d\right)=100$
$\Rightarrow a_{100}-a_{100}^{\prime}=\left(a-a^{\prime}\right)=100$
So, difference does not depend on number of terms. Thus, $a_{1000}-a_{1000}^{\prime}=100=a_{100}-a_{100}^{\prime}$
Hence, the difference between their 1000 th terms is 100.
$a_n=a+(n-1) d \text { and }$
$a_n^{\prime}=a^{\prime}+(n-1) d, \text { respectively }$
Now 100 th term of 1 st AP will be given by: $a_{100}=a+(100-1) d=a+99 d$
100th term of second AP will be given by: $a^{\prime}{ }_{100}=a+(100-1) d=a^{\prime}+99 d$
Now given, $a _{100}- a ^{\prime}{ }_{100}=( a +99 d)-\left( a ^{\prime}+99 d\right)=100$
$\Rightarrow a_{100}-a_{100}^{\prime}=\left(a-a^{\prime}\right)=100$
So, difference does not depend on number of terms. Thus, $a_{1000}-a_{1000}^{\prime}=100=a_{100}-a_{100}^{\prime}$
Hence, the difference between their 1000 th terms is 100.
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