Question

Find the middle term of the sequence formed by all three-digit numbers which leave a remainder when divided by $7.$ Also find sum of all numbers on both sides of the middle terms separately.

Answer 1

A.P $103,110,117,\dots \dots ..,999$

$a=103$

$d=110-103=7$

n be the number of terms

$a_n=999$

$103+(n-1)7=999$

$(n-1).7=999-103$

$\Rightarrow (n-1).7=896$

$\Rightarrow n-1=896/7$

$\Rightarrow n-1=128$

$\Rightarrow n=128+1$

$\Rightarrow n=129$

Middle term $={\frac{n+1}{2}}^{th}$ term $=\frac{129+1}{2}={65}^{th}$ term.

$=a+6d=103+64\times 7=551$

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

$S_{64}=\frac{64}{2}[2.103+63.7]$

$\Rightarrow S_{64}=20704$

$S_{129}=\frac{129}{2}[2.103+128.7]$

$\Rightarrow S_{129}=71079$

$S_{129}-(S_{64}+551)=71079-(20704+551)=49824$.