Find the middle term of the sequence formed by all three digits numbers which leave a remainder 3 when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

The numbers which leave 3 as remainder when divided by 4 will be in the form 3q + 4, for any integer q.

Hence, three digit numbers which leave 3 as remainder when divided by 4 are:

103, 107, 111, …, 199

Clearly above sequence is an AP, with

First term, a = 103

Common difference, d = a₂ – a = 107 – 103 = 4

Let the number of terms be ‘n’.

We know, nth term of an AP is

a_n = a + (n – 1)d

we obtain last term = 199

⇒ a_n = 199

⇒ a + (n – 1)d = 199

⇒ 103 + (n – 1)4 = 199

⇒ 4(n – 1) = 96

⇒ n – 1 = 16

⇒ n = 17

∴ Number of terms = 17

Hence, middle term =

and a₉ = a + 8d

= 103 + 8(4)

= 135

Hence, middle term of the sequence is 135

Also, we know

We know, sum of ‘n’ terms of an AP is

Now, sum of terms left to the middle term(9^th term)S_left = S₈

= 4(206 + 28)

= 936

Now, sum of terms right to the middle (9^th term) S_right = S₁₈ – S₉

= 2466 – 1071

= 1395