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

Solution A

① Defining the two sequences.

Let the first arithmetic sequence be $\{a_{n}\}$.

Let the second arithmetic sequence be $\{b_{n}\}$.

We can define two A.P by

• $d$, the common difference
• $n$, the term number
• $a$, the first term of the sequence $\{a_{n}\}$
• $b$, the first term of the sequence $\{b_{n}\}$

The two sequences can be defined by

$\implies a_{n}=a+(n-1)d$

$\implies b_{n}=b+(n-1)d$

② Given condition.

According to the given condition,

$\implies |a_{100}-b_{100}|=100$

Putting the defined sequences,

$\implies |a+99d-b-99d|=100$

$\implies |a-b|=100$

And we test $|a_{n}-b_{n}|$,

$=|a+(n-1)d-b-(n-1)d|$

$=|a-b|=\boxed{100}$

So, the difference of the sequences is always 100, regardless of the term number.

Answer 2

Given :-

Two APs have the same common difference. The difference between their 100th terms is

100,

To Find :-

difference between their 1000th terms?

Solution :-

We know that

aₙ = a + (n - 1)d

a₁₀₀ = a + (100 - 1)d

a₁₀₀ = a + (99)d

a₁₀₀ = a + 99d

a'₁₀₀ = a' + (n - 1)d'

a'₁₀₀ = a' + (100 - 1)d'

a'₁₀₀ = a' + 99d'

Now

a + 99d - a' - 99d = 100

a + a' = 100