Question

Find the sum of all 3-digit numbers, which leave the re-
mainder 1, when divided by 4.​

Answer 1

Step-by-step explanation:

The first 3-digit number that leaves a remainder of 2 when divided by 3 = 101.

The next numbers being 104, 107, 110, 113,……………

The last 3-digit number that leaves a remainder of 2 when divided by 3 = 998.

So, practically, these numbers form an AP series 101, 104, 107, 110, 113,…….., 998.

First term of this AP series = 101.

Common difference = 3.

Final (n-th) term of the series = 998.

Therefore, 101 + (n - 1)*3 = 998; which leads to n = 300.

So, sum of the numbers in this AP series = 300*(101 + 998) / 2 = 164850.

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Answer 2

Answer:

101, 105, ....997

$a_{n} = a + (n-1)d$

997 = 101 + (n-1)4

224 = n - 1

225 = n

$S_{n} = \frac{n}{2}(a+l)$

$S_{225} = \frac{225}{2} (101 + 997)$

$S_{225} = \frac{225}{2} (1098)\\S_{225} = 225(549)\\ = 123525$

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