Arithmetic Sequence & Progression

Concept: nth term of an arithmetic progression,

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.

Question: How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

  1. 128
  2. 142
  3. 143
  4. 141
  5. 129

Explanatory Answer

Video explanation will be added soon

Find the first and last term of the series

The smallest 3-digit positive integer that leaves a remainder of 5 when divided by 7 is 103.

The largest 3-digit positive integer that leaves a remainder of 5 when divided by 7 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. The common difference of the sequence is 7.

Compute the number of terms

In an A.P, the last term 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

So, n - 1 = 128 or n = 129

Compute the number of terms

In an A.P, the last term 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

So, n - 1 = 128 or n = 129

Choice E is the correct answer.

Are you targeting Q-51 in GMAT Quant? Make it a reality!

Comprehensive Online classes for GMAT Math. 20 topics.
Focused preparation for the hard-to-crack eggs in the GMAT basket!

Try it Free Now
online gmat classes

Next Only Sundays GMAT batch starts Sep 23, 2017. Call +91 95000 48484.

Sun 6:30 AM to 1:30 PM @ Nungambakkam, Chennai Start Now