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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.
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
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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