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 the numbers on both sides of the middle term.

Answer 1

List of $3$ digit no leaves remainder of $3$ divided by $4$

$103,107,111,115...999$

The above list of $AP;a=103$

$d=4$

$a_n=999$

$a+(n-1)d=999$

$103+(n-1)4=999$

$n=225$

Middle term

$=(n+12{)}^{th}term$

$={113}^{th}$ term

$=a+112d$

$=551$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{112}{2}[2\times 103+111\times 4]$

$=36400$

Sum of all numbers $=\frac{225}{2}[2\times 103+224\times 4]$

$=123975$

sum after middle term $=S_{225}-(S_{112}+551)$

$=123975-(36400+551)$

$=87024$