Find the middle term of the sequence formed by three digit number which leave the remainder 3 and when divided by 4 als o find the all number on both sides the middle term seprately
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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.
middle term = (n+12)th term = 65th term = a + 64d = 103 + 64×7 = 551We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]Now, S64 = 642[2×103 + 63×7] = 20704Sum of all terms before middle term = S64 = 20704S129 = 1292[2×103+128×7] = 71079Now, sum of term
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.
middle term = (n+12)th term = 65th term = a + 64d = 103 + 64×7 = 551We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]Now, S64 = 642[2×103 + 63×7] = 20704Sum of all terms before middle term = S64 = 20704S129 = 1292[2×103+128×7] = 71079Now, sum of term
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