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Sia ? 6 years, 2 months 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) {tex}\times{/tex} 5
5 (n – 1) = 997 – 102 = 895
n-1={tex}\frac {885}{5}{/tex}
(n – 1) = 179
n = 179 +1
n = 180
Sum of all three digit numbers which leaves remainder 2 when divided by 5
Sn ={tex}\frac{n}{2}{/tex} [a+l]
Sn= {tex}\frac{180}{2}{/tex} [ 102+ 997]
Sn= 90 {tex}\times{/tex} 1099
Sn= 98910
Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910
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