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 middle term.

Answer 1

The three digit number when divided by 7 divided reminder 5 are.

103, 110, 117,----- 999.

This forms an A.P with common difference = 7 and first term = 103.

Now $a_n=a+(n-1)d$

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

$\frac{896}{7}=n-1$ $\Rightarrow n=128+1=129.$

The middle term of sequence is ${65}^{th}$term.

and it is

$a_{65}=103+64\times 7=103+448$

$[a_{65}=551].$

Now sum of term before ${65}^{th}$ term

$s_n=\frac{n}{2}[2a+(n-1\left)d\right]$ $=\frac{64}{2}[2\times 103+63\times 7]$

$s_n=32[206+441]=20,704.$

sum of term after ${65}^{th}$term

$S_n=\frac{64}{2}[2\times 558+63\times 7]=49,824.$

$\therefore$[S before middle term = 20,704] & [S after middle term = 49824].

and middle term is $[a_{65}=551]$