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The first 4 terms of a sequence are 2, 6, 18, and 54.

Each term after that is three times the preceding term.

If the sum of the 49th, 50th and 51st terms of this sequence is expressed as k⋅3^{49}, then k = ?

Click Read more to see the answer, solution, discussion...

Answer: 26/3

Suggested Solution

The first three terms can be written as

2(3^{0}), 2(3^{1}) and 2(3^{2}). (***)

In general, the nth term is 2(3^{n-1}).

The sum of the 3 desired terms would then be 2(3^{48}) + 2(3^{49}) +2(3^{50}). Factoring out 3^{49}, we obtain 3^{49}(2/3 + 2 + 6) = (26/3)(3^{49}), so k = 26/3.

Comments

(1) Too hard for SATs? Similar (but slightly easier) problems have appeared on the test.

(2) Could students use the "Make it simpler strategy" here to reduce the problem to the sum of just the first three terms? But this is the essence of geometric sequences (or exponential functions):

From (***) above, this sum would be

2(3^{0}) + 2(3^{1}) + 2(3^{2}) = 2 + 6 + 18 = 3(2/3 + 2 +6) = 3(26/3). The coefficient 26/3 would be the same for any three consecutive terms! Is this concept/technique worth developing?

## Friday, April 17, 2009

### Classic Exponent Challenge for SATs, Algebra 2, Math Contests...

Posted by Dave Marain at 7:37 PM

Labels: advanced algebra, exponents, geometric sequence, math contest problems, more, SAT-type problems

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## 4 comments:

Dave,

first, let me thank you for continuing to post these sorts of problems.

But then, an almost political (not careful) comment: How could a problem like this ever be accessible to a kid whose algebra foundation is weak? who sat through an "algebra lite"?

Yet in New York, we now have an algebra course ("Integrated Algebra") without much depth. I don't know what's happening in California, but they have algebra for everyone at the same age (8th grade) - I can imagine an lcd course in weaker districts... and in average districts. Constructivist middle school programs in the suburbs are labeled "algebra" and the kids are hustled into geometry the following year.

Now, in some districts, the course, the teaching, they will go deeper than the test. But not in most. And if "algebra lite" is the order of the day, your question will only separate the top of the top... much more basic questions will screen out the middle.

Jonathan

Jonathan

Jonathan--

You are certainly describing a disturbing trend of which I am unfortunately aware.

Yes, the difficulty level of the problem I posed is considered 'Level 5' (most discriminating) by the College Board. However, simpler versions of this type of question could be introduced as early as prealgebra or Algebra 1:

2+4+8 = 2(1+2+4) = 7(2^1)

4+8+16 = 4(1+2+4) = 7(2^2)

8+16+32 = 8(1+2+4) = 7(2^3)

Concrete numerical patterns are usually the basis for further generalizations using algebra and rules of exponents. I would also argue that geometric sequences can be introduced in middle school as a way of motivating or reinforcing the ideas of exponents.

Unfortunately, no one is writing these kinds of curricular materials and most would assume that such "challenges" are appropriate only for the stronger accelerated student.

The issues surrounding the "Algebra for All" movement are of critical importance. Some 8th, 9th or even 10th graders simply are NOT READY to succeed in a traditional Algebra I program because they do not have the necessary prerequisite skills. IMO,

these students need intensive re-training in arithmetic and prealgebra during the school year and during the summer to be ready for the next level. To address this reality, many educators have instead proposed an alternative to a standard algebra program -- a program that is less traditional and more applied. A program where you and I would ask, "Where's the meat?" There are no simple answers here...Good Morning -

This question would be considered too "hard" for the SAT, not because of some inherent difficulty level but because of the number of steps, number of "realizations" and time required to finish.

It's a valuable question for students who have more than 30 seconds to consider it.

The make it simpler strategy is a valuable one - and definitely one that should be explored with students.

Curmudgeon,

You're right that this problem may be too hard for the SATs but a few points..;.

(1) A similar question

hasappeared on the test. Whenever I develop my own questions based on actual College Board problems, I intentionally raise the level somewhat. I'm a great believer in preparing students by challenging them with questions that are slightly "above" the test. However, the actual question involved the same concepts and was quite challenging. The difference was that the College Board problem was multiple choice and more susceptible to "plugging in" and testing the choices.(2) Your comment on "number of steps" is interesting. See my reply to you re the percent question posted recently. In fact, some SAT questions are not testing insight as much as the student's ability to work accurately through a series of steps -- as many as FIVE steps!

Again, I will repeat the sample problem posted on my other comment to demonstrate this (very close to a "real" problem):

.What is the perimeter of a square whose area is the same as that of a circle whose circumference is 6This question requires considerable algebraic skill and strong organizational ability.

(3) Finally, I'm hoping my readers recognize that my primary intent in writing "sample" questions is to stimulate thought and discussion about the inherent mathematical content (concepts, procedures, skills). The test-taking strategies are also very important but are secondary to the main purpose. Writing these questions and devising high-quality distractors is difficult. I know I'm leaving myself open to criticism but that's ok!

Curmudgeon (and I'm sure your moniker belies a much softer kinder side!), keep this dialog going. Your insights and experience are valuable both for students and teachers who visit here.

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