What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
A. 897 B. 1,64,850 C. 1,64,749 D. 1,49,700 Answer: Option BShow Answer
Solution(By Apex Team)
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ….
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.
Sum of an AP =
$\left(\Large\frac{\text { First Term }+\text { Last Term }}{2}\right) \times \text { Number of Terms }$
We know that in an A.P., the nth term $a_{n}=a_{1}+(n-1) \times d$
In this case, therefore, 998 = 101 + (n – 1) × 3
i.e., 897 = (n – 1) × 3
Therefore, n – 1 = 299
Or n = 300
Sum of the AP will therefore, be
$\begin{array}{l}=\left(\Large\frac{101+998}{2}\right)\times300\\
=1,64,850\end{array}$
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