Find the middle term of the sequence formed by all three digit no. which leaves a remainder 3 , when divided by 4. Also , find the sum of all no's. on both sides of the middle terms separately
Answers
The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :
103 , 107 , 111 , 115 , .... 999
The above list is in AP with first term, a = 103 and common difference, d = 4
Let n be the number of terms in the AP.
Now, an = 999
103 + ( n - 1 ) 4 = 999
103 + 4n - 4 = 999
4n + 99 = 999
4n = 900
n = 225
Since, the number of terms is odd, so there will be only one middle term.
middle term = (n+12)th term = 113th term = a + 112d = 103 + 112×4 = 551
We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]
Now, Sum = 112/2[2×103 + 111×4] = 36400={S112}
Sum of all terms before middle term = 36400
sum of all numbers= 225/2[2×103+224×4] = 123975={S225}
Now, sum of terms after middle term = S225− (S112+551) = 123975−(36400+551) = 87024