Question

find the middle term of the sequence formed by all 3 digit numbers which leave a remainder 5 when divided by 7 also find the sum of all numbers after the middle term​

Answer 1

what u try to say bro or sis.

Answer 2

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

a = 103  d = 7

n be the number of terms

= 999

103 + (n - 1)7 = 999

103 + 7n - 7 = 999

7n + 96 = 999

7n = 903

n = 129

Middle term =  term =  = a + 64d = 103 + 64 x 7 = 551

=  [2a + (n - 1)d]

=  [2 x 103 + 63 x 7] = 20704

=  [2 x 103 + 128 x 7] = 71079

- ( + 551) = 71079 - (20704 + 551) = 49824