Question

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

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 AP.
Now, a_n = 999
103 + (n-1)7=999
103+7 n -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 = $S_{64}=20704S_{129}=\frac{129}{2}[2\times 103+128\times 7]=71079$
Now, sum of terms after middle term =$S_{129}-(S_{64}+551)=71079-(20704+551)=49824$