The GMAT Math practice question given below is a sequences and series question based on Arithmetic Progressions about finding number of terms of an Arithmetic sequence.
How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
The correct choice is (E) and the correct answer is 129
The smallest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 103.
The largest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 999.
The series of numbers that satisfy the condition that the number should leave a remainder of 5 when divided by 7 is an A.P (arithmetic progression) with the first term being 103 and the last term being 999 having a common difference of 7.
We know that in an A.P, 'l' the last term is given by l = a + (n - 1) * d, where 'a' is the first term, 'n' is the number of terms of the series and 'd' is the common difference.
Therefore, 999 = 103 + (n - 1) * 7
Or 999 - 103 = (n - 1) * 7
Or 896 = (n - 1) * 7
Or n - 1 = 128
Or n = 129
More Questions in Sequences & Series : AP GP