Find the sum of all 3-digit numbers Which leave the remainder 2 when divided by 5.
Answers
Answered by
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
==================================================================================
Hope this will help you....
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
==================================================================================
Hope this will help you....
Answered by
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
Similar questions