Question

Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately.

Answer 1

The list of 3 digit number that leaves a remainder of 5 when divided by 7 is:
103, 110, 117, ......, 999
The above list is in AP with first term, a = 103 and common difference, d = 7
Let n be the number of terms in the A.P.
Now, $a_n=999$
103 + (n - 1)7 = 999
103 + 7n - 7 = 999
7n + 96 = 999
7n = 903
n = 129
Since, the number of terms is odd, so there will be only one middle term.
middle term $={\left(\frac{n+1}{2}\right)}^{th}term={65}^{th}term=a+64d=103+64\times 7=551$
We know that, sum of first n terms of an AP is,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Now, $S_{64}=\frac{64}{2}[2\times 103+63\times 7]=20704$
Sum of all terms before middle term $=S_{64}=20704$
$S_{129}=\frac{129}{2}[2\times 103+128\times 7]=71079$
Now, sum of terms after middle term $=S_{129}-(S_{64}+551)=71079-(20704+551)=49824$