Math, asked by BrainlyHelper, 1 year ago

Find the sum of all 3-digit numbers Which leave the remainder 2 when divided by 5.

Answers

Answered by nikitasingh79
291
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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Hope this will help you....
Answered by zakirmohd002
6

Answer:

Hope it's may useful for you

Step-by-step explanation:

Three digit numbers which leave theremainder when divided by 5 are 102,107,112,...997

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