Question

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?​

Answer 1

Answer:

Let first term of 1st AP = a

Let first term of 2nd AP = a’

It is given that their common difference is same.

Let their common difference be d.

It is given that difference between their 100th terms is 100.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a + (100 – 1) d – [a’ + (100 – 1) d] = a + 99d – a’ – 99d = 100

⇒ a – a’ = 100 … (1)

We want to find difference between their 1000th terms which means we want to calculate:

a + (1000 – 1) d – [a’ + (1000 – 1) d] = a + 999d – a’ – 999d = a – a’

Putting equation (1) in the above equation,

a + (1000 – 1) d – [a’ + (1000 – 1) d] = a + 999d – a’ + 999d = a – a’ = 100

Therefore, difference between their 1000th terms would be equal to 100.

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Answer 2

Answer:

Let a and A be the first and second term of the APs and 'd' be the common difference.

According to question,

$\\ \qquad \sf \: a_{100} - A_{100} = 100 \\ \\ \\ \implies \sf \: a + 99d \: - \: ( \: A + 99d) = 100 \\ \\ \\ \implies \sf \: a + \cancel{99d} \: - A + \cancel{99d} = 100 \\ \\ \\ \implies \sf \: a - A = 100 \: \: \qquad \: - \: - \: - \: \: \: \: (i)$

Now, the common difference between their 1000th term -

$\\ \\ \sf a_{1000} - A_{1000} = a + \cancel{999d} - A - \cancel{999d} \\ \\ \\ \implies \sf \: a - A \\ \\ \\ \implies \sf \: 100 \: \: \: \qquad \: \: ( \: from \: i \: ) \\ \\ \\ \large{ \boxed{ \sf{a_{1000} - A_{1000} = 100}}}$

•°• Therefore, the common difference between their 1000th term is 100.