Find the middle term of the sequence formed by all three digit number which leave a remainder 5 when divisible by 7. Also find the sum of all numbers on both sides of the middle term separately
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The list of 3 digit number that leaves a remainder of 5 when divided by 7 is :
103, 110, 117, ........., 999
The above list is in AP with first term, a = 103 and common difference, d = 7
Let n be the number of terms in the AP.
Now, an = 999
103 + ( n - 1 ) 7= 999
103 + 7n - 7 = 999
7n + 96 = 999
7n = 903
n = 129
Since, the number of terms is odd, so there will be only one middle term.
HOPE IT HELPS YOU!!!
The list of 3 digit number that leaves a remainder of 5 when divided by 7 is :
103, 110, 117, ........., 999
The above list is in AP with first term, a = 103 and common difference, d = 7
Let n be the number of terms in the AP.
Now, an = 999
103 + ( n - 1 ) 7= 999
103 + 7n - 7 = 999
7n + 96 = 999
7n = 903
n = 129
Since, the number of terms is odd, so there will be only one middle term.
HOPE IT HELPS YOU!!!
harsh1234567892:
bro this is half answer
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