Find the sum of all three digit numbers which leave the remainder 3 when divided by 5
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Here, According to the above condition,
First 3-digit number will be 103.
Next, 108,113.....
We can see
103,108,113,118......,998.
So, it is an AP.
Now, we know that
t = a+(n-1)d
998 = 103+(n-1)5
895 = (n-1)5
n-1 = 179
n = 180.
Now, Sn = n/2 (a+l) = 180/2 (103+998)
= 90*1101 = 99090.
Hope you will find it helpful....
MARK IT AS BRAINLIEST.
First 3-digit number will be 103.
Next, 108,113.....
We can see
103,108,113,118......,998.
So, it is an AP.
Now, we know that
t = a+(n-1)d
998 = 103+(n-1)5
895 = (n-1)5
n-1 = 179
n = 180.
Now, Sn = n/2 (a+l) = 180/2 (103+998)
= 90*1101 = 99090.
Hope you will find it helpful....
MARK IT AS BRAINLIEST.
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