Question

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.

Answer 1

The sequence is 103, 110, ......, 999

∴ 999 = 103 + (n – 1) × 7 

⇒ n = 129

Therefore (129+1)/2 = 65th term is the middle term

Middle term = 103 + (64) × 7 = 551

Sum of first 64 terms = 32 [206 + 63 × 7] = 20704

Sum of last 64 terms= 32 [1116 + 63 × 7] = 49824