Question

what is the answer of this question​

Answer 1

The List of 3 digit number that leaves a remainder of 3 when divided by 4 is :-

103, 107, 111 ,115 ... 999  

As this is in the form of an AP

a = 103

d = 4

An = 999

103 + (n - 1) 4 = 999

103 + 4n - 4 = 999

4n + 99 = 999

4n = 999 - 99

4n = 900

n = 225  

Middle term = (n+12)th term (113th Term)

a + 112d = 103 + (113-1) × 4

a + 112d = 551  

Sn = n/2 [ 2a + (n−1) d]  

$Sn_{1}$ = 112/2 [2 × 103 + 111 × 4]

Sn1 = 36400  

Sum of all terms before middle term  = 36400

Sn2 = 225/2 [2 × 103 + 224 × 4]

= 123975

Now, sum of terms after middle term

Sn2 − (Sn1 + 551)

= 123975 − (36400 + 551)

= 87024