Find the middle term of the sequence formed by all three digit numbers which leave a remainder 5when divided by 7
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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
Hope I helped you! :)
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
Hope I helped you! :)
Answered by
1
No : Is this ans no it is not a answer
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