<p>a, 2a + 1, 2(2a+1)+1, ...</p>
<p>In the sequence above, the first term is a and each term after the first is found by doubling the preceding term and then adding 1. If a is a positive integer and if exactly 4 terms in the sequence are less than 1,000, which of the following could be value of a?</p>
<p>A) 32
B) 49
C)61
D)109
E)125</p>
<p>So the fifth term must be 1000 or greater.</p>
<p>That term is: a, 2a + 1, 2(2a+1) + 1 = 4a + 3, 2(4a + 3) + 1 = 8a + 7, 2(8a + 7) + 1 = 16a + 15</p>
<p>Solve: 16a + 15 = 1000 and you get a = 61 + 9/16. So the answer is (C)</p>
<p>… answer is not C</p>
<p>This means that the fifth term MUST be greater than 1000. Fogcity has already worked out the algebraic way to determine the 5th term correctly, so I will just use 16a + 15 as the 5th term. So therefore:</p>
<p>16a + 15 > 1000
16a > 985
a > 61.5625 or 61 and 9/16</p>
<p>However, since it must be GREATER than 61.5625, that rules out C, which is only 61. Therefore, the answer is D 109.</p>
<p>Conversely a different way to work this out is through plug and chug because this type of question lends itself very easily to a guess and check. All you have to do is choose a number and plug them in manually because the term in question is only the 5th term. (If you’re asked about the 100th term then obviously not a good idea to be guessing and checking) So pick the middle number, 61 and try it out. It should take a max of 30 seconds to solve this problem. 61 gives you 991 as the 5th term, which then you know you just missed it by a hair and you can try 109 which will give you 879 as the 4th term and 1759 as the 5th term.</p>
<p>Here’s how I would do it:</p>
<p>Start with choice (C) as your first guess. Then a=61 and the sequence becomes</p>
<p>61, 123, 247, 495, 991. So there are 5 terms less than 1000, and we can eliminate (A), (B) and (C).</p>
<p>Let’s try (D) next. Then a = 109 and the sequence becomes</p>
<p>109, 219, 439, 879, 1759</p>
<p>So (D) is the answer.</p>
<p>First step of EVERY math question is to look at the answers to determine one important thing: which strategy you’re gonna use.</p>
<p>1) Plug (in the answers…basically test them out)
2) Choose (your own numbers for variables)</p>
<p>MOST of the time if you see actual numbers in the answer choices, like you see here, you’re gonna use Strategy #1: Plug in the Answers (PITA)…a pain in the ass</p>
<p>MOST of the time if you see variables anywhere in the answer choices, you’re going to use Strategy #2: Choose your own numbers (for the variables)</p>
<p>If you realize one strategy doesn’t work, simply switch over to the other strategy.</p>
<p>So start with choice C because it’s the middle number (they organize choices in ascending order), so if you guess with the middle number you may not have to check all choices…you’ll know if you need a bigger or smaller guess if C doesn’t work.</p>
<p>So PLUG in choice C (where a = 61). That gives you:
a = 61
2a + 1 = 123
and so on…just like DrSteve said.
Keep going until you go over 1,000. Count how many are less than 1,000…well, there’s five numbers less than 1,000 here, so choice C is wrong.</p>
<p>You want to go up to choice D because C (being a smaller number) produced too many numbers less than 1,000. To decrease the number of #s less than 1,000, you need to start with a bigger number like choice D.</p>
<p>When you plug in choice D, you get: 109, 219, 439, 839, 1759…four of those are less than 1,000…so you have your winner!</p>
<p>My preferred method is plugging numbers, so I would go with DrSteve’s solution.</p>
<p>Nevertheless, I like exploring alternatives; here are the two of them.</p>
<p>1.
Each term after the first one is roughly double the preceding one (the original sequence is growing a little (in the beginning) faster than that), so the fourth term is about 2^3 times greater than the first one.
If the fourth term is 999, then the first one is about 999/8 = 124.875.
We need to start with a bit smaller number, thus (D) 109.</p>
<ol>
<li>
Just for entertainment purposes (don’t try this on the test).
In your TI-8X enter in Y= editor
Y1 = x
Y2 = 2Y1 + 1
Y3 = 2Y2 + 1
Y4 = 2Y3 + 1
Y5 = 2Y4 + 1
Go to TABLE and plug x=61: Y4=495, Y5=991.
Now plug x=109: Y4=879, Y5=1759. Done.</li>
</ol>
<p>If you are using a graphing calculator, here’s another useful tip: the enter key is really a “repeat last instruction” key.</p>
<p>For example, press 1 and hit enter.</p>
<p>Then, press “+1” and hit enter again. You will see:ans(1)+1=2.</p>
<p>Now, keep hitting enter and each time, it will take the last answer and increase it by 1.</p>
<p>So why is this useful? Well, in this case, we need to try different starting values and apply the double and add 1 rule. So here goes:</p>
<p>Enter “61” and press enter.</p>
<p>Then type “x2+1” and hit enter again. (That’s the times symbol, not an x)</p>
<p>Your screen will show: ans(1)*2+1 </p>
<p>and when you hit enter you will see: 123</p>
<p>Now keep hitting enter and watch what happens…</p>