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Question

Find the middle term of sequence formed by all three digits number which leaves a reminder 5 when divided by 7.also find the sum of all number on both sides of the middle term separately.

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Solution

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


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