Question

Find the middle term of the sequence formed by all the three digit numbers which leave a remainder 3 when divided by 4. Also find the sum of all the numbers on the left side of the middle term.

Answer 1

Sum of the term on right of middle term=87024...

I hope it's help you..

Answer 2

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.


now , u follow the picture.