Sequences and Series Practice Questions

First, let us simplify the problem by rephrasing the question. Since any even number must be divisible by \(2\), any even multiple of \(15\) must be divisible by \(2\) and by \(15\), or in other words, must be divisible by \(30\). As a result, finding the sum of even multiples of \(15\) is equivalent to finding the sum of multiples of \(30\).<br><br>

By observation, the first multiple of \(30\) greater than \(295\) will be equal to \(300\) and the last multiple of \(30\) smaller than \(615\) will be equal to \(600\). Thus, since there are no multiples of \(30\) between \(295\) and \(299\) and between \(601\) and \(615\), finding the sum of all multiples of \(30\) between \(295\) and \(615\), inclusive, is equivalent to finding the sum of all multiples of \(30\) between \(300\) and \(600\), inclusive.<br><br>

Therefore, we can rephrase the question: "What is the greatest prime factor of the sum of all multiples of \(30\) between \(300\) and \(600\), inclusive?"<br><br>

The sum of a set = (the mean of the set) × (the number of terms in the set)<br><br>

Since \(300\) is the \(10\)th multiple of \(30\), and \(600\) is the \(20\)th multiple of \(30\), we need to count all multiples of \(30\) between the \(10\)th and the \(20\)th multiples of \(30\), inclusive.<br><br>

There are \(11\) terms in the set: \(20\)th – \(10\)th + \(1 = 10 + 1 = 11\)<br><br>

The mean of the set = (the first term + the last term) divided by \(2\): \(\frac{300 + 600}{2} = 450\)<br><br>

\(k =\) the sum of this set \(= 450 \times 11\)<br><br>

Note, that since we need to find the greatest prime factor of \(k\), we do not need to compute the actual value of \(k\), but can simply break the product of \(450\) and \(11\) into its prime factors:<br><br>

\(k = 450 \times 11 = 2 \times 3 \times 3 \times 5 \times 5 \times 11\)<br><br>

Therefore, the largest prime factor of \(k\) is \(11\).<br><br>

The correct answer is C.

For sequence S, any value \( S_n \) equals 6n. The problem asks for the sum of all multiples of 6 between 78 (\( S_{13} \)) and 168 (\( S_{28} \)), inclusive.<br><br>The sum can be found by identifying the median of the set and multiplying by the number of terms. Because this set includes an even number of terms, the median equals the average of the two middle terms, \( S_{20} \) and \( S_{21} \), or (120 + 126)/2 = 123.<br><br>Given that there are 16 terms in the set, the answer is 16 × 123 = 1,968.<br><br>The correct answer is D.

Let the five consecutive even integers be represented by x, x + 2, x + 4, x + 6, and x +8. Thus,<br>the second, third, and fourth integers are x + 2, x + 4, and x + 6. Since the sum of these three<br>integers is 132, it follows that<br><br>3x + 12 = 132, so<br>3x = 120, and \( x = 40 \).<br><br>The first integer in the sequence is 40 and the last integer in the sequence is x + 8, or 48.<br>The sum of 40 and 48 is 88<br>The correct answer is C.