Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. 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 7 will be in the form 7q + 5, for any integer q.

Hence, three digit numbers which leave 5 as remainder when divided by 7 are:

103, 110, 117, …, 999

Clearly above sequence is an AP, with

The first term, a = 103

Common difference, d = a₂ – a₁
= 110 – 103
= 7

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 = 999

⇒ a_n = 999

⇒ 103 + (n – 1)7 = 194

⇒ 103 + (n – 1)7 = 999

⇒ 7(n – 1) = 896

⇒ n – 1 = 128

⇒ n = 129

∴ Number of terms = 129
Middle term will be:

middle term is 65th term.
Middle term is :
103 + 64(7) = 103 + 448
= 551

We know, sum of ‘n’ terms of an AP is

Now, sum of terms left to the middle term (64 terms), S_left = S₆₄

S₆₄ = 32(206 + 63(7))

     = 32(206 + 441)
     = 32(647)
    = 20704

Now, the sum of terms right to the middle term(last 64 terms)

As 64th term is 551.
now 65th term will be 558.

= 32(1116 + 63(7))
= 32(1116 + 441)
= 32(1557)
= 49824