Question

Find the sum of all 3 digit numbers which leave the remainder 3 when divided by 5

Answer 1

first 3 digit number leave a remainder 3 when divided by 5 = 103

last 3 digit number leave a remainder 3 when divided by 5 = 998

common difference = 5

total terms

998 = 103 + (n-1)5

998 - 103 = (n-1)5

895 = (n-1)5

895÷5 = n-1

179 = n-1

n = 180

Sn =

$= \frac{n}{2} (2a + (n - 1)d) \\ = \frac{180}{2} (2 \times 103 + (180 - 1)5) \\ = 90(206 + (179 \times 5)) \\ = 90(206 + 895) \\ = 90 \times 1101 \\ = 99090$

Answer 2

Answer:

Step-by-step explanation:

Solution :-

We observe that 103 is the is the 1st three digit number and 998 is the last three digit number which leaves the remainder 3 when divided by 5.

Therefore, AP = 103, 108, 113,.....,998

Here, a = 103, d = 5, nth term = 998

a(n) = a + (n - 1)d

⇒ 998 = 103 + (n - 1)5

⇒ 998 - 103 + 5 = 5n

⇒ 900 = 5n

⇒ n = 900/5

⇒ n = 180

Then, S(n) = n/2(2a + (n - 1)d)

⇒ S(n) = 180/2(2 × 103 + (180 - 1)5)

⇒ S(n) = 90(206 + 895)

⇒ S(n) = 90 × 1101

⇒ S(n) = 99090

Hence, the sum of all 3 digit number 99090.