Question Video: Finding the Number of Geometric Means Inserted between Two Numbers under a Given Condition Mathematics

Find the number of geometric means inserted between 82 and 1,312 given the sum of the last two means equals twice the sum of the first two means.

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Video Transcript

Find the number of geometric means inserted between 82 and 1,312 given the sum of the last two means equals twice the sum of the first two means.

If we know that geometric means have been inserted between 82 and 1,312, then 82 and 1,312 are part of a geometric sequence. We can identify 82 as the first term in the sequence, π‘Ž sub one. We know that 1,312 is the final term in the sequence, but we don’t know how many terms there are. So we’ll label it as π‘Ž sub 𝑁.

The general term of a geometric sequence with the first term π‘Ž sub one and a ratio of π‘Ÿ is π‘Ž sub 𝑁 equals π‘Ž sub one times π‘Ÿ to the 𝑁 minus one power. This means we can write the equation π‘Ž sub 𝑁 minus two plus π‘Ž sub 𝑁 minus one equals two times π‘Ž sub two plus π‘Ž sub three. The left side of our equation, it represents the sum of the last two means, and the right side we have two times the sum of the first two means.

We want to rewrite all of these values in terms of π‘Ž sub one and π‘Ÿ. π‘Ž sub two, that’s the first mean after our initial value, is equal to 82 times π‘Ÿ to the two minus one power. And π‘Ÿ to the first power is just π‘Ÿ. The second mean in the sequence would be 82 times π‘Ÿ squared. π‘Ž sub 𝑁 minus two is a little bit more difficult to find. It’s equal to 82 times π‘Ÿ to the 𝑁 minus two minus one power, which simplifies to 82 times π‘Ÿ to the 𝑁 minus three power. And then π‘Ž sub 𝑁 minus one equals 82 times π‘Ÿ to the 𝑁 minus two power.

Then we plug these values back into the equation that we initially wrote. On the left, we can factor out the term 82 times π‘Ÿ to the 𝑁 minus three power. Remember the power rule that tells us π‘Ž to the π‘₯-power times π‘Ž to the 𝑦-power equals π‘Ž to the π‘₯ plus 𝑦 power. If you multiply 82 times π‘Ÿ to the 𝑁 minus three power times π‘Ÿ, you’re multiplying by π‘Ÿ to the first power. And 𝑁 minus three plus one equals 𝑁 minus two. This is what happens when we take out a factor of π‘Ÿ to the 𝑁 minus three power from π‘Ÿ to the 𝑁 minus two power.

On the right-hand side, the factoring is a little bit easier. We take out a term of 82π‘Ÿ, and that leaves us with one plus π‘Ÿ.

Remember that our π‘Ÿ-value is a ratio. And if you have a negative ratio in your geometric sequence, then the terms will alternate negative, positive, negative, positive. We can eliminate negative one as an option for our π‘Ÿ as that would make the geometric sequence 82, negative 82, 82, negative 82, and so on. If we divide through by one plus π‘Ÿ on both sides, we’ll end up with 82π‘Ÿ to the 𝑁 minus three power equals 164π‘Ÿ. We have the 164 because we’ve multiplied 82 by two on the right side. From there, we can divide both sides of the equation by 82π‘Ÿ to the first power.

Again, we’ll remember a power rule π‘Ž to the π‘₯-power over π‘Ž to the 𝑦-power equals π‘Ž to the π‘₯ minus 𝑦 power. On the left, we’ll have 82 divided by 82 equals one, and π‘Ÿ to the 𝑁 minus three power divided by π‘Ÿ becomes π‘Ÿ to the 𝑁 minus four power, and on the right, 164π‘Ÿ divided by 82π‘Ÿ equals two.

At this point, we have an equation with two unknowns. And this means to solve, we’ll need another equation. We know what π‘Ž sub 𝑁 equals, 1,312. We also know that π‘Ž sub 𝑁 can be rewritten as 82 times π‘Ÿ to the 𝑁 minus one power. If we plug in 1,312 here and then divide both sides by 82, we get 16 equals π‘Ÿ to the 𝑁 minus one power.

What we wanna try to do now is rewrite π‘Ÿ to the 𝑁 minus four power in terms of π‘Ÿ to the 𝑁 minus one power. We can do that by saying π‘Ÿ to the 𝑁 minus one power times π‘Ÿ to the negative three power, again using our power rules. We know that π‘Ÿ to the 𝑁 minus one power equals 16. So we can substitute 16 in here.

How do we deal with this π‘Ÿ to the negative three power? We can rearrange this, but it’ll take a few steps. First, we rewrite π‘Ÿ to the negative three power as one over π‘Ÿ cubed. From there, we multiply both sides of the equation by π‘Ÿ cubed and then divide by two. We have eight equals π‘Ÿ cubed. The cube root of eight is two, and the cube root of π‘Ÿ cubed equals π‘Ÿ. Remember that π‘Ÿ-variable represents the ratio in this geometric sequence. This means the second term in the sequence will be 82 times two, 164. The third term will be 164 times two, 328. The fourth term will be 656. And 656 times two equals 1,312.

This means that we know our π‘Ž sub 𝑁 is π‘Ž sub five. 1,312 is the last term in the sequence, and it’s the fifth term. But when we count the geometric means, we do not count the first term or the last term, as these are not means, which means the number of geometric means inserted between 82 and 1,312 under these conditions is three.

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