Video: Finding the Value of a Term of a Geometric Sequence Under a Given Condition

Find the value of π‘Žβ‚‚ in a geometric sequence given 6π‘Žβ‚‚ + π‘Žβ‚ƒ = 16π‘Žβ‚, π‘Žβ‚β‚€ = 62 and all terms are positive.

06:25

Video Transcript

Find the value of π‘Ž sub two in geometric sequence given six π‘Ž sub two plus π‘Ž sub three is equal to 16 π‘Ž sub one, π‘Ž sub 10 is equal to 62, and all terms are positive.

So the first thing to look at is the notation we’ve got. So we’ve got π‘Ž sub two and π‘Ž sub three, etc. And what this means is the term. So, for instance, π‘Ž sub two is the second term, π‘Ž sub three will be the third term, etc. Okay, we’re also looking at geometric sequence. And what a geometric sequence is, is a sequence where there is a common ratio between each term. And with the geometric sequence, we have a form to help us find any term within our geometric sequence. And that is that π‘Ž sub 𝑛 is equal to π‘Ž sub one multiplied by π‘Ÿ to the power of 𝑛 minus one. So what I’m gonna do is use this in conjunction with the information we’ve got to help us solve the problem.

Well, first of all, we’re gonna take look at the tenth term. And that is equal to 62. And then if we sub this into our general form, we’re gonna get the tenth term is equal to the first term multiplied by π‘Ÿ to the power of 10 minus one. So therefore, it’s gonna be the tenth term is equal to the first term multiplied by π‘Ÿ to the power of nine, where π‘Ÿ is a common ratio. As we said, being a geometric sequence is a common ratio between terms. So therefore, what we can write is the first term multiplied by the common ratio to the power of nine is equal to 62. Okay, great, so that’s used one piece of information. Now, let’s go on and use the other bits of information that we have.

We were told that six multiplied by the second term plus the third term is equal to 16 multiplied by the first term. So let’s first of all find the second and third terms in terms of the first term. So our second term is gonna be equal to the first term multiplied by π‘Ÿ to the power of two minus one, which is gonna give us a second term is equal to the first term multiplied by the common ratio. And the third term is gonna be equal to the first term multiplied by the common ratio squared. Ok, great, so how’s this gonna be useful? Well, what we can do is we can substitute these in in place of our second and third terms in our equation that we’ve got. So when we do that, we’re gonna get six π‘Ž sub one π‘Ÿ plus π‘Ž sub one π‘Ÿ squared is equal to 16 π‘Ž sub one.

Now, if we take a look at left-hand side, we can see that we’ve got a common factor because π‘Ž sub one or our first term is a common factor in both of our terms. So therefore, what we can do is we can rewrite our equation as π‘Ž sub one multiplied by six π‘Ÿ plus π‘Ÿ squared is equal to 16π‘Ž sub one. Well, therefore, what we can do is divide through by π‘Ž sub one. And when we do that, we’re gonna get six π‘Ÿ plus π‘Ÿ squared is equal to 16. So now, so that I can set up a quadratic equation is equal to zero so that we can solve, what I’m gonna do is subtract 16 from each side. So we get π‘Ÿ squared plus six π‘Ÿ minus 16 is equal to zero. So now, what we’re looking for, because we’re gonna factor our quadratic, is two factors whose sum is positive six and whose product is negative 16. Well, our two factors are gonna be positive eight and negative two. And that’s because if we multiply positive eight and negative two, we get negative 16. And eight minus two is six.

Okay, great, so we’ve got our factors. So now, what we do is we find out what our values of π‘Ÿ will gonna be. Well, we get π‘Ÿ is equal to negative eight or two. We got that because to give a value of zero on the right-hand side of the equation, one of our parentheses is gonna have to be equal to zero. So if we had π‘Ÿ plus eight is equal to zero, then π‘Ÿ will be negative eight. And if we had π‘Ÿ minus two equals zero, then π‘Ÿ would have to be equal to two. However, are we gonna accept both these values for π‘Ÿ? Well, no, because we’re told that all terms of our geometric sequence are positive. So therefore, the common ratio cannot be negative. So therefore, we can disregard the common ratio, which is negative eight. So we now know that π‘Ÿ, our common ratio, is equal to two.

So now, we found our common ratio. What we’ll look and find is the value of the second term. But we can’t quite find it because we still need to find out what the first term is. That’s π‘Ž sub one. Well, in order to do that, what we’re going to do is substitute in π‘Ÿ equals two into the equation that we had for π‘Ÿ sub one multiplied by π‘Ÿ to the power of nine equals 62. We found that because we knew what the tenth term was. Well, we now know that the first term multiplied by two to the power of nine is equal to 62.

So therefore, we know that the first term, π‘Ž sub one, multiplied by 512 is equal to 62. So now, what we’re gonna do is divide through by 512. So we get the first term. π‘Ž sub one is equal to 62 over 512. What I’m gonna do now is just simplify the fraction on the right-hand side. And to do that, what I’ve done is I’ve divided both the numerator and denominator by two because that’s the factor of both numbers. And what I get is π‘Ž sub one is equal to 31 over 256.

Okay, great, so we now have the first term. We now have the common ratio. So what we can do is substitute this in to find the value of this second term. So what we’re gonna do, now that we found both the first term and the common ratio, is substitute this in. And when we do this, what we get is the second term is equal to 31 over 256 multiplied by two, which is gonna give us a second term is equal to 62 over 256. We could’ve just halve the value on the denominator. That would’ve given us the final answer. But I’ve done this by multiplying the numerator by two. So we get 62 over 256. Again, we can simplify this. Once again, we can do this by dividing the numerator and denominator by two. And once we do this, we get that the second term is equal to 31 over 128.

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