Question Video: Finding the Geometric Sequence given Its First and Last Terms and the Sum of All the Terms | Nagwa Question Video: Finding the Geometric Sequence given Its First and Last Terms and the Sum of All the Terms | Nagwa

# Question Video: Finding the Geometric Sequence given Its First and Last Terms and the Sum of All the Terms Mathematics • Second Year of Secondary School

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Find the geometric sequence given the first term is 324, the last term is 4, and the sum of all the terms is 484.

03:58

### Video Transcript

Find the geometric sequence, given the first term is 324, the last term is four, and the sum of all the terms is 484.

At first glance, it might feel like we don’t have enough information to answer this question. But we’re dealing with a geometric sequence. So let’s recall what we do know about a geometric sequence. The 𝑛th term of a geometric sequence, 𝑎 sub 𝑛, is given by 𝑎 one times 𝑟 to the power of 𝑛 minus one, where 𝑎 sub one is the first term in the sequence and 𝑟 is the common ratio. It’s the number we multiply each term by to get the next term.

Similarly, we’ve been given some information about the sum of the terms in the sequence. And so we recall the sum of the first 𝑛 terms of the geometric sequence, with a first term 𝑎 sub one and a common ratio 𝑟, is 𝑎 sub one times one minus 𝑟 to the 𝑛th power over one minus 𝑟.

Now we’re told the first term in our sequence is 324. In other words, 𝑎 sub one is 324. We know the last term is four. But we don’t actually know how many terms are in our sequence. So we can write 𝑎 sub 𝑛 to be equal to four. And then, of course, we know the sum of all our terms to be equal to 484.

Let’s substitute everything we have into each of our formulae. Our first formula, the one for the 𝑛th term, becomes four equals 324 times 𝑟 to the power of 𝑛 minus one. We’re going to divide through by 324. Four divided by 324 is one over 81. So we find that 𝑟 to the power of 𝑛 minus one is equal to one over 81.

In our other formula, we know that the sum of the first 𝑛 terms is 484. So we say that 484 is equal to 324 times one minus 𝑟 to the 𝑛th power all over one minus 𝑟. We’re going to multiply through by one minus 𝑟 here. And that gets rid of the fraction for us. So we have 484 times one minus 𝑟 equals 324 times one minus 𝑟 to the 𝑛th power.

Let’s distribute each pair of parentheses. On the left-hand side, multiplying both one and negative 𝑟 by 484 gives us 484 minus 484𝑟. And on the right, we get 324 minus 324 times 𝑟 to the 𝑛th power. And now we have an equation in terms of 𝑟 to the power of 𝑛 minus one and then one that has 𝑟 to the power of 𝑛 and 𝑟.

So we’re going to do something a little bit strange with this first equation. We’re going to multiply through by 𝑟. And we’ll see why we do that in a moment. On the left-hand side, when we do, we get 𝑟 over 81, nothing groundbreaking. And on the right, we get 𝑟 to the power of 𝑛 minus one times 𝑟 or 𝑟 to the power of one. And of course, we know 𝑟 is the same number. When we multiply two numbers with the same base, we can simply add their exponents. So this becomes 𝑟 to the power 𝑛 minus one plus one, which is just 𝑟 to the power of 𝑛. And that’s great because we can now replace 𝑟 to the power of 𝑛 in our second equation with 𝑟 over 81. And we’ll have an equation purely in terms of 𝑟. We get 484 minus 484𝑟 equals 324 minus 324𝑟 over 81.

Now in fact, 324 divided by 81 is four. So this becomes four 𝑟. We’re going to rearrange this equation by subtracting 324 from both sides and adding 484𝑟 to both sides. That tells us that 160 is equal to 480𝑟. And we can now solve for 𝑟 by dividing through by 480. 160 divided by 480 is simply one-third.

So we found the common ratio. 𝑟 is equal to one-third. And that’s great because we know that, to get each term in our sequence, we multiply the first term repeatedly by 𝑟. So if the first term is 324, the second term is 324 times one-third, which is 108. The third term is either 324 times a third squared or 108 times a third, which is 36. We do that two more times and we actually end up with the final term in our sequence, the number four.

So we’ve calculated 𝑟 to be one-third. And actually, in doing so, we’ve been able to deduce the fact that 𝑛 is equal to five. There are five terms in our sequence. They are 324, 108, 36, 12, and four.

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