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?

- 128
- 142
- 143
- 141
- 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.

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