###
**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 B**

## Show 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
$\begin{array}{l}=\left(\Large\frac{101+998}{2}\right)\times300\\
=1,64,850\end{array}$

## Related Questions On Progressions

### How many terms are there in 20, 25, 30 . . . . . . 140?

A. 22B. 25

C. 23

D. 24

### Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

A. 5B. 6

C. 4

D. 3

### Find the 15th term of the sequence 20, 15, 10 . . .

A. -45B. -55

C. -50

D. 0

### The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

A. 600B. 765

C. 640

D. 680