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