Question

Find the sum of all three digit numbers each of which leaves the remainder2, when divided by 3​

Answer 1

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Step-by-step explanation:
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
Sn = n/2 [a+l]
Sn= 180/2 [ 102+ 997]
Sn= 90 × 1099
Sn= 98910