find no of all three digit natural no which are divisible by 7 and leaves remainder 3
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first three digit number is 100
if 100 divides by 7 then remainder is 2
so 101 is the first number who when divides by 7 remaining 3
next number must be greater by 7
so it is 108
so series is
101 , 108 , 115 ,.........
last three digit number is 999 when divides by 7 remaining 5
so 997 is the number when divides by 7 remaining 3
so series is
101 , 108 , 115 ,......... ,997
here a = 101
d = 7
tn = 997
n = ?
tn = a + (n-1) d
997 = 101 + ( n - 1 ) × 7
997 - 101 = ( n - 1 ) × 7
( n - 1 ) × 7 = 896
n - 1 = 896 / 7
n - 1 = 128
n = 128 + 1
n = 129
so 129 numbers
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