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
Solution(By Examveda 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 =
$$\left( {\frac{{{\text{First Term}} + {\text{Last Term}}}}{2}} \right) \times $$ $${\text{Number}}\,{\text{of}}\,{\text{Terms}}$$
We know that in an A.P., the nth term an = a1 + (n - 1) × 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
$$\eqalign{ & = \frac{{101 + 998}}{2} \times 300 \cr & = 1,64,850 \cr} $$
Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
Join The Discussion