Breaking Down a GMATPrep Sequences Problem
This week, were going to tackle a harder GMATPrep problem solving question from the topic of Sequences (a subset of Algebra).
Lets start with the problem. Set your timer for 2 minutes. and GO!
*In the infinite sequence [pmath]a_1[/pmath], [pmath]a_2[/pmath], [pmath]a_3[/pmath], , [pmath]a_n[/pmath], , each term after the first is equal to twice the previous term. If [pmath]a_5[/pmath] [pmath]a_2[/pmath] = 12, what is the value of [pmath]a_1[/pmath]?(A) 4
(B) 24/7
(C) 2
(D) 12/7
(E) 6/7
So, we need to know some terms here. A sequence is a collection of numbers in a certain order. The order is determined by a particular rule. For instance, if I tell you that a sequence starts with the number 1 and the rule is to add 2 to each successive number, then whats the second number in the sequence? And the third?
The second number is 3 and the third number is 5. The sequence (so far!) is {1, 3, 5}. If I tell you that its an infinite sequence, then the sequence continues forever, using the same pattern or rule. If I tell you the sequence has 10 terms, then you would continue the pattern until there were 10 numbers in the set and then stop.
In the problem above were told that we have an infinite sequence and were also given the rule: the second term is double the first time; the third term is double the second term; and so on. If our first term is 1, then the second term is 2 and the third term is 4.
Hmm. They dont tell us the first term, though they ask us to find the first term. Instead, they give us some information about two other terms, [pmath]a_5[/pmath] and [pmath]a_2[/pmath].
There are two main ways to approach this one, depending upon your comfort level with sequences. The first will take a bit longer but doesnt require you to understand sequences super-well. The second is faster, but you really need to know how to handle sequences in order to use it.
Heres our first approach (the longer one):
[pmath]a_1[/pmath] = ? (this is what I want to find)
[pmath]a_2[/pmath] = 2[pmath]a_1[/pmath]
[pmath]a_3[/pmath]= 2[pmath]a_2[/pmath]
[pmath]a_4[/pmath] =2[pmath]a_3[/pmath]
[pmath]a_5[/pmath] =2[pmath]a_4[/pmath]
Were also given this equation: [pmath]a_5[/pmath] [pmath]a_2[/pmath] = 12. Can you use the above equations to derive a second equation that also contains those two terms?
Start substituting! Begin with the equation containing [pmath]a_2[/pmath] that is closest to [pmath]a_5[/pmath]. (Two equations contain [pmath]a_2[/pmath]; which one should you use?)
If [pmath]a_3[/pmath] = 2[pmath]a_2[/pmath], I can plug the right-hand side of that equation into this one: [pmath]a_4[/pmath] =2[pmath]a_3[/pmath]. I get:
[pmath]a_4[/pmath] =2(2[pmath]a_2[/pmath])
Substitute again with the next equation, [pmath]a_5[/pmath] =2[pmath]a_4[/pmath], to get:
[pmath]a_5[/pmath] =2[2(2[pmath]a_2[/pmath])]
[pmath]a_5[/pmath] =8[pmath]a_2[/pmath]
Voil! Weve got another equation with the two variables we want, [pmath]a_5[/pmath] and [pmath]a_2[/pmath]!
How should we combine the last equation above with the equation given in the problem? Substitution again.
First, rearrange this equation: [pmath]a_5[/pmath] [pmath]a_2[/pmath] = 12 [pmath]a_5[/pmath] = 12 + [pmath]a_2[/pmath]
Then substitute into the other:
12 + [pmath]a_2[/pmath] = 8[pmath]a_2[/pmath]
12 = 7[pmath]a_2[/pmath]
12/7 = [pmath]a_2[/pmath]
Great! Weve got our answer!
Wait. No, we dont. We just solved for [pmath]a_2[/pmath], and that is in the answer choices but they asked us for [pmath]a_1[/pmath]. Careful! Whats the relationship between [pmath]a_2[/pmath]and [pmath]a_1[/pmath]? The same as all the others: [pmath]a_2[/pmath] is twice as big as [pmath]a_1[/pmath]. If I know [pmath]a_2[/pmath], I have to divide by 2 (or multiple by ) in order to find [pmath]a_1[/pmath].
(12/7) * (1/2) = 12/14 = 6/7
The correct answer is E.
How can we do this more efficiently? If you really understand the above (which is why I presented it first!) then you can learn a shortcut that will let you save time on this question.
Sequences can be defined with direct formulas or recursive formulas. A direct formula is one in which the rule is provided as a function of n, the place in which the term occurs in the sequence. For example, given this formula:
[pmath]a_n[/pmath]= 3n + 2
n refers to the placement of the term in the sequence, and [pmath]a_n[/pmath] refers to the value of the term. For instance, the 1st term is where n=1, and the value of that term is 3(1) + 2 = 5. The 2nd term is where n=2, and the value of that term is 3(2) + 2 = 8. Whats the third term?
A recursive formula can define the same rule, but it is written a bit differently. A recursive formula provides the rule as a function of a previous term or terms in the sequence. For example, given this formula:
[pmath]a_n[/pmath] = [pmath]a_{n-1}[/pmath] + 3
[pmath]a_n[/pmath] refers to the value of any particular term, and that term is the sum of the value of the previous term in the sequence, [pmath]a_{n-1}[/pmath], and 3. In order to solve for the values in this sequence, you actually need to know one value in the sequence. For instance, the first term is where n=1, but I cant solve for [pmath]a_1[/pmath] because I dont know what [pmath]a_0[/pmath] is. If Im told, however, that [pmath]a_1[/pmath] = 5, then I can solve for [pmath]a_2[/pmath] (and all other terms). [pmath]a_2[/pmath] = [pmath]a_1[/pmath] + 3 = 5 + 3 = 8. Whats the third term in this sequence?
Generally speaking, most people find direct formulas easier to use than recursive formulas.
There are two main categories of sequence rules that everyone first learns when we learn sequences: arithmetic and geometric. Arithmetic sequences add a constant to each successive term. For example, if Im adding 3 to each successive term, then 3 is my constant and this is an arithmetic sequence (and this is exactly the example given in the discussion of direct and recursive formulas above!).
Geometric sequences multiply each successive term by a constant. For example, if I have the sequence 1, 3, 9, 27, 81, Im multiplying each successive term by 3.
Look back at our problem again. Do we have a geometric or an arithmetic sequence? (Note: we may not have either one.)
We do! Its a geometric sequence. Each successive number doubles, or is multiplied by 2. Great. The annoying thing about the way the problem is presented right now: its written in terms of a recursive formula. We dont want to take all those steps we took in solving the first time. The great news is that we dont have to, now that we know we have a geometric sequence.
The first term is [pmath]a_1[/pmath]. The rule is always going to be to multiply each successive term by 2. If my first term is [pmath]a_1[/pmath], the second term is 2[pmath]a_1[/pmath]. Whats the third term? 4[pmath]a_1[/pmath]. And then 8[pmath]a_1[/pmath] and 16[pmath]a_1[/pmath] and so on.
We still want to know something about the second and fifth terms, because those are the terms used in the formula given in the problem. The second term, [pmath]a_2[/pmath], is 2[pmath]a_1[/pmath] and the 5th term, [pmath]a_5[/pmath], is 16[pmath]a_1[/pmath]. How did we get from the 2nd term to the 5th term? We multiplied by 8. (Is this starting to ring a bell?)
So the relationship between [pmath]a_2[/pmath] and [pmath]a_5[/pmath] is this:
[pmath]a_5[/pmath] = 8[pmath]a_2[/pmath]
Great! We just skipped half the math from our first solution method. We could continue with the rest of the solution from our first method.
It gets even better though. We don't even have to figure out the relationship between the two. We have these three formulas: [pmath]a_2[/pmath]=2[pmath]a_1[/pmath], [pmath]a_5[/pmath]=16[pmath]a_1[/pmath], and [pmath]a_5[/pmath] [pmath]a_2[/pmath] = 12. Plug and chug!
16[pmath]a_1[/pmath] 2[pmath]a_1[/pmath] = 12
14[pmath]a_1[/pmath] = 12
[pmath]a_1[/pmath] = 12/14
[pmath]a_1[/pmath] = 6/7
And were done! Note that, with this method, we solve directly for [pmath]a_1[/pmath]. Thats faster and now we cant fall into the trap of picking answer D (the value for [pmath]a_2[/pmath]), because were not even calculating it!
Key Takeaways for Solving Hard Sequence Problems:
- Direct sequences / rules are often easier to handle than recursive sequences. If you can translate a rule given in recursive terms into a direct rule instead, do so.
- For geometric sequences, the rule is always to multiply each successive term by a constant number (note: that constant can be anything, including a fraction or negative!). If you need to solve across multiple terms (as we actually did in the problem above), then take the constant and raise it to the power of the number of terms you need to move.
- For arithmetic sequences, the rule is always to add a constant number to each successive term (note: that constant can be anything, including a negative!). If you need to solve across multiple terms (similar to the above problem), then multiply the constant by the number of terms you need to move.
- When youre studying, try to find ways to solve for the desired term without first having to solve for other terms, which might show up in the answers as traps. (On the real test, go with whatever occurs to you right away dont spend a ton of time trying to come up with a more elegant solution. Ditto when youre practicing problems under timed conditions. When youre going over problems, however, take all the time you need to find better methods!)
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.