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Find the sum of all three digit numbers which leave remainder 2 when divided by 5. (A) 98910 (B) 9820(C) 9830 (D) 9840

Find the sum of all three digit numbers which leave remainder 2 when divided by 5. (A) 98910 (B) 9820(C) 9830 (D) 9840

Grade:10

1 Answers

Arun
25750 Points
4 years ago
Three digit numbers which leave the remainder 2 when divided by 5 are 102, 107, 112,117......, 997.
102, 107, 112,......, 997 is an A.P

In this A.P 
a (first term)= 102
d (Common difference)= 5

I(last term ) = 997
 
l= an = a + (n – 1) d
997= 102 + (n – 1) × 5
5 (n – 1) = 997 – 102 = 895

n-1= 885/5
(n – 1) = 179
n = 179 +1

n = 180
Sum of all three digit numbers which leaves remainder 2 when divided by 5

Sn = n/2 [a+l]

Sn= 180/2 [ 102+ 997]

Sn= 90 × 1099

Sn= 98910

Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910

 
 

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