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 numbers on both sides of the middle term separately ?

Answer 1

We know that the smallest three digit number is 100. Now, 100 is exactly divisible by 4, 101 leaves a remainder 1 when divided by 4, 102 leaves a remainder 2 when divided by 4 and 103 leaves a remainder 3 when divided by 4.
So, next term after 103 that will leave a remainder 3 when divided by 4 is 107 (103 + 4).
Now, all three digit-numbers that leave a remainder 3 when divided by 4 are 103, 107, 111, 115 , ..., 999, which forms an AP.
We know
n^th term of the AP = T_n = a + (n − 1)d
Here, a = 103 and d = 4.
So,

103 + (n − 1) × 4 = 999
⇒ 103 + 4n − 4 = 999
⇒ 4n = 999 − 99 = 900
⇒ n = 225, which is odd
∴ Middle term of the sequence = \[\left( \frac{225 + 1}{2} \right)th\] term of the sequence

                                                  = 113th term of the sequence
                                                  = 103 + (113 − 1) × 4
                                                  = 103 + 112 × 4
                                                  = 551

Thus, the middle term of the sequence is 551.

So, the AP can be rewritten as follows:

103, 107, 111, 115 , ..., 547, 551, 555, ..., 999
There are 112 terms before and after the middle term 551 in the AP.
Now,
Sum of all numbers before the middle term \[= \frac{112}{2}\left( 103 + 547 \right)\]

\[= 56 \times 650\]

\[ = 36, 400\]

Thus, the sum of all numbers before the middle term is 36,400.
Sum of all numbers after the middle term \[= \frac{112}{2}\left( 555 + 999 \right)\]

\[= 56 \times 1554\]

\[ = 87, 024\]

Thus, the sum of all numbers after the middle term is 87,024.