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 terms separately

Answer 1

hello dear...............

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

a = 103  d = 7

n be the number of terms

 an= 999

103 + (n - 1)7 = 999

103 + 7n - 7 = 999

7n + 96 = 999

7n = 903

n = 129

Middle term = n+1/2
term = 65th= a + 64d = 103 + 64 x 7 = 551

 sn= n/2 [2a + (n - 1)d]

 s64= 64/2 [2 x 103 + 63 x 7]
 s64= 20704

s129 = 129/2 [2 x 103 + 128 x 7]
 s129= 71079

 = s129-(s64 + 551) = 71079 - (20704 + 551)
= 49824

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hope this would help you
@Rohit