Question

find the middle term of the sequence formed by all three digit numbers which leaves remainder 5 when divided by 7 also find the sum of all numbers on both sides of the middle term separately

Answer 1

A.P: 103, 110, 117, ...., 999

a = 103  d = 7

n be the number of terms

a_{n}an​ = 999

103 + (n - 1)7 = 999

103 + 7n - 7 = 999

7n + 96 = 999

7n = 903

n = 129

Middle term = \frac{n+1}{2} ^{th}2n+1​th term = 65^{th}65th = a + 64d = 103 + 64 x 7 = 551



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