Question 4 the day : April 17, 2006
The question for the day is a sample practice problem in Sequences and series - AP and summation of terms of an AP. This question appeared in CAT 2003.

Question
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

1. 897
2. 164,850
3. 164,749
4. 149,700

Correct Choice is (2) and Correct Answer is 164,850.

Explanatory Answer

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 =

We know that in an A.P., the nth term a_n = 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 = 164,850

Question on finding sum of terms of a GP, Problem on finding a particular term of an AP, given sum of two other terms, Summation of AP having negative common difference, Finding terms of an AP, given sum and product of some of its terms, Sum of series of numbers in arithmetic progression, A CAT 2003 question to find sum of terms of an AP

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