Find the middle term of the sequence formed by all three digit numbers which leave a reminder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately.
Answers
Answered by
5
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) = 48924
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) = 48924
Pankhi5:
Can you be a little more specific in the last part of the answer.
Sry i can not type now my hand is raining I have got ingure other wise I would have helped u
Okay, Fine.
Sry for inconvenience
i can understand
Similar questions