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Video: Geometric Sequences with Algebraic Terms

Tim Burnham

We explain how to tackle geometric sequence questions in which you are given algebraic expressions as terms (e.g., find possible values of x and a general formula for the nth term of a geometric sequence with 1st 3 terms 2, x - 2, and x + 10).

09:15

Video Transcript

In this video, we’re gonna use your skills of algebraic manipulation in order to solve some geometric sequence problems.

First, let’s do a little warm up with this question. The fourth and fifth terms in a geometric sequence are one hundred and twenty-five and six hundred and twenty-five respectively. Find the first, second, third, and tenth terms of the sequence. Right, first let’s just define some terms. I’m gonna use π‘Ž one to represent the first term in the sequence π‘Ž 𝑛 to represent the 𝑛th term, so π‘Ž two is the second term, π‘Ž three is the third term, and so on. And then I’m gonna use the letter π‘Ÿ to represent the common ratio between terms. So we’ve been given the fourth and fifth terms, so π‘Ž four is a hundred and twenty-five and π‘Ž five is six hundred and twenty-five. And remember, to get one term, we need to multiply the previous term by the common ratio. So π‘Ž five for example is π‘Ž four times the common ratio. so that means in our case that six hundred and twenty-five is one hundred and twenty-five times π‘Ÿ. And dividing both sides of that equation by a hundred and twenty-five, I get six hundred and twenty-five over one hundred and twenty-five is equal to π‘Ÿ. In other words, in this case, π‘Ÿ is equal to five.

So just remember then that to work out the value of π‘Ÿ, the common ratio, we take the value of one term and we divide it by the value of the immediately preceding term. So, so far we’ve worked out the value of the common ratio is five. So to work out the value of π‘Ž four we just take π‘Ž three and multiply it by π‘Ÿ. Now we know the value of π‘Ž four; we know the value of π‘Ÿ. So we can now work out the value of π‘Ž three, the third term. hundred twenty-five equals π‘Ž three times five. So dividing both sides by five means that the third term is equal to twenty-five. And the third time is equal to the second term times π‘Ÿ. Well now we know the third term, and we know the value of π‘Ÿ, so we can work out the value of the second term. twenty-five equals π‘Ž two times five. So, again we can divide both sides by five. And we now know that the value of the second term is five. Also, the value of the second term is the value of the first term times the common ratio. Well the value of the second term is five; the value of the common ratio is five. So I’ve got five is equal to the first term times five. Dividing both sides by five tells me that the first term has a value of one. So I’ve worked out the value of the first, second, and third terms. To work out the value of the tenth, I could just roll this procedure forward but I’m gonna work out the value of the 𝑛th term, so the formula for the 𝑛th term, and then apply that to the value ten.

Now the general formula for the 𝑛th term of a geometric sequence is that the 𝑛th term is equal to the first term times the common ratio to the power of 𝑛 minus one. Well, I know that the first term has a value of one. I know that the common ratio is five. So I can now complete this formula. 𝑛th term is one times five to the power 𝑛 minus one. Well one times that, I don’t really need to write one times that, so I can make that little bit more efficient. So my 𝑛th term formula is π‘Ž 𝑛, the 𝑛th term, is equal to five to the power of 𝑛 minus one. Now I can use that formula to calculate the value of the tenth term, so I just put in value of 𝑛 equals ten. So my tenth term is going to be five to the power of ten minus one. So that’s five to the power of nine. And when I put that into my calculator, I get one million nine hundred and fifty-three thousand one hundred and twenty-five. So just checking back to the question to make sure we’ve answered it. We had to find the value of the first, second, third, and tenth terms of the sequence. Well the first, second, and third terms are one, five, and twenty-five, and the tenth term as we just said is one million nine hundred and fifty-three thousand one hundred and twenty-five.

Okay now we’re warmed up; let’s do another question that actually genuinely involves a bit more algebra. A geometric sequence that consists of only positive terms has the first three terms two, π‘₯ minus two, and π‘₯ plus ten. Find the possible values of π‘₯ and the general formula for the 𝑛th term of the sequence. So we’re told that the first term is two, the second term is π‘₯ minus two, and the third term is π‘₯ plus ten. Now we weren’t told what the common ratio was, but we do know that the common ratio π‘Ÿ is simply one term divided by the immediately preceding term. So it could be π‘Ž two divided by π‘Ž one or π‘Ž three divided by a two for example. And the consequence of that is that π‘Ž two divided by π‘Ž one must give us the same answer as π‘Ž three divided by π‘Ž two. Now we’ve got expressions for π‘Ž one, π‘Ž two and π‘Ž three, so let’s put them into one great big equation. So that means (π‘₯ minus two) over two is equal to (π‘₯ plus ten) over (π‘₯ minus two). Now if I multiplied both sides of that equation by two, then the two cancels from the left-hand side just giving me π‘₯ minus two, and then I’ve got an extra two on the numerator on the right-hand side. Now I’m going to multiply both sides my equation by π‘₯ minus two to try to eliminate the denominator from the right-hand side. So on the left-hand side, I had my π‘₯ minus two and then I multiply that by π‘₯ minus two. And on the right-hand side, the π‘₯ minus two that I am multiplying by cancels the denominator, so it just leaves me with two lots of π‘₯ plus ten.

So we’re just making a little bit a space for ourselves. We’ve got (π‘₯ minus two) times (π‘₯ minus two) is equal to two times (π‘₯ plus ten). So I’m gonna multiply those out on the left-hand side, multiplying each term in the second bracket by each term in the first bracket and distributing the two across the brackets or parentheses on the right-hand side. So on the left, I’ve got π‘₯ squared minus two π‘₯ minus another two π‘₯ plus four; and on the right-hand side, I got two times π‘₯ is two π‘₯ plus two times ten is twenty. So here, I’ve got minus two π‘₯ minus another two π‘₯, so I can combine those to make minus four π‘₯. So I’ve got a quadratic expression, so what I’m gonna do is try and solve that. So I’ve got to make that equal to zero, so I’m gonna subtract two π‘₯ from both sides and then subtract twenty, giving me a quadratic equals to zero which hopefully I’ll be able to factor and then find out the values of π‘₯. So first of all let’s subtract two π‘₯. So on the left-hand side, we’ve got π‘₯ squared minus four π‘₯ minus another two π‘₯ is minus six π‘₯, and then we’re still got our plus four. And on the right hand side, we’re taking away that two π‘₯ which just leaves us with twenty. So now let’s subtract twenty. So on the left hand side, I’ve got π‘₯ squared minus six π‘₯ plus four take away twenty is minus sixteen, negative sixteen. And on the right-hand side, if I takeaway twenty, I’ve just got nothing. So now I’ve got a quadratic which is equal to zero. If I can factor that quadratic, I can find out the values of π‘₯. And luckily, it’s quite easy to factor: (π‘₯ plus two) times (π‘₯ minus eight). So those two things multiplied together are equal to zero, and the only way you can get an answer of zero if you multiply two things together is if one of them is zero. So either π‘₯ plus two is zero or π‘₯ minus eight is zero. And if π‘₯ plus two is equal to zero, then π‘₯ will be equal to negative two. And to make π‘₯ minus eight equal to zero, π‘₯ would need to be equal to eight. So we’ve got two possible values of π‘₯.

Now remember, the question said that the geometric sequence consists of only positive terms, so these terms have to be positive. We also know the first term is two, the second term is π‘₯ minus two and the third term is π‘₯ plus ten. So if π‘₯ was equal to negative two, the second term would be, negative two take away two, be negative four. So that would break the rules; so that can’t actually be a solution. So the only value of π‘₯ that satisfies the instructions in the question, if you like, is π‘₯ equals eight. And if π‘₯ equals eight, I can now write out the first three terms of my sequence properly. Well π‘Ž one is still two, π‘Ž two is π‘₯ minus two so that’s eight minus two, which is six. And π‘Ž three is π‘₯ plus ten, so that’s eight plus ten is eighteen. So the first three terms are two, six, and eighteen. And I can work out the common ratio π‘Ÿ just by doing the second term divided by the first terms, so that’s six divided by two. So the common ratio is three. And the general formula for the 𝑛th term of any sequences, π‘Ž 𝑛, the 𝑛th, term is equal to the first term π‘Ž one times the common ratio π‘Ÿ to the power of (𝑛 minus one). Now we’ve got values for π‘Ž one and π‘Ÿ. My 𝑛th term formula becomes the 𝑛th term, π‘Ž 𝑛, is equal to two times three to the power of (𝑛 minus one). And strictly you don’t need to put the parentheses around the three, but it just makes it absolutely clear that it’s only the three that is to the power of 𝑛 minus one and not the two as well. So we’re just going back and checking we’ve answered everything. Find the possible values of π‘₯. Well there is only one possible value of π‘₯ if the sequence is only gonna consist of positive terms. And we wanted the general formula for the 𝑛th term of the sequence and we’ve got that here. Okay! Looks like we’re done.