Question

Common difference of two arithmetic progressions is same. If the difference of

their 99th terms is 99, then what will be the difference of their 999th term?​

Answer 1

Answer:-

Given:

Common difference of two AP's is same.

Let the first AP be a1 , a1 + 2d....

and 2nd be a2 , a2 + 2d....

And also given that,

Difference between their 99th terms is 99.

We know that,

nth term of an AP = a + (n - 1)d

Hence,

99th term of first AP - 99th term of 2nd AP = 99

→ a1 + (99 - 1)d - [ a2 + (99 - 1)d ] = 99

→ a1 + 98d - [a2 + 98d] = 99

→ a1 + 98d - a2 - 98d = 99

→ a1 - a2 = 99

Hence, difference of their first terms is 99.

Now,

We have to find the difference of their 999th terms.

→ 999th term of first AP - 999th term of 2nd AP = ?

→ a1 + (999 - 1)d - [ a2 + (999 - 1)d ] = ?

→ a1 + 998d - (a2 + 998d) = ?

→ a1 + 998d - a2 - 998d = ?

→ a1 - a2 = ?

Putting the values from equation (1) we get,

→ Difference between their 999th terms = 99

Hence, the difference between the 999th terms of two AP's is 99.