Question

find the middle term of the sequence formed by all three digit numbers which leave a remainder 3, when divided by 4. also find the sum of all numbers on both the sides of the middle term

Answer 1

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
Sum of all terms before middle term = 36400
sum of all numbers= 225/2[2×103+224×4] = 123975

Now, sum of terms after middle term = S225 − (S112+551) = 123975−(36400+551) = 87024