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Eleventh grader Kausar asks, "I had this question several times but i do not know how to do it. You are given the 5th and 7th terms of a geometric sequence. Can you determine the 29th term without finding the general term?"
Hi Kausar, the answer to your question is, "Yes, you can." Having said that, I'd like to add that if you'd been asked for the 28th term, or the 30th term, the answer would be "No, you can't." Let's take a look at why.
Let's say the fifth term is 32, and the seventh term is 8. Then we can reason as follows:
"The sixth term is the fifth term times the common ratio, and the seventh term is the sixth term times the common ratio. Therefore, the seventh term is the fifth term times the common ratio squared."
Algebraically, this reasoning looks like this:
a6 = a5r; a7 = a6r; a7 = (a5r)r = a5r2. Thus, 8 = 32r2, which leads to r2 = 1/4.
Now, without bothering to get the general term, we ask, "How many times would I have to multiply the seventh term by r2 to get the 29th term? And the answer is: (29 - 7)/2 = 11.
So a29 = a7r11 = 8(1/4)11, which is approximately 1.9x10-6.
So why does this work for the 29th term, but not the 28th? Because (28 - 7)/2 is not an integer, so we can't get there by multiplying by r2. We'd have to find r, and multiply by r (28 - 7) times.
So why can't we do that? Well...because the equation r2 = 1/4 doesn't have just one solution; it has two. r = 1/2 or r = -1/2. So you have two possible values for the 28th term.
I occasionally get hired to write problems for math competititions, and one of the "tricks" I'll occasionally pull on students is to give a sequence like this, where students are likely to end up with an equation like r2 = 1/4, and then forget that this has two solutions instead of just one. It's amazing how often a simple trick like that will trip students up!
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