Lesson Video: Sum of a Finite Geometric Sequence Mathematics • 10th Grade

In this video, we will learn how to calculate the sum of the terms in a geometric sequence with a finite number of terms.

17:31

Video Transcript

In this video, we will learn how to calculate the sum of the terms in a geometric sequence with a finite number of terms. We will begin by recalling what we mean by a finite geometric sequence. In a geometric sequence, each term is found by multiplying the previous term by a constant, for example, the sequence two, six, 18, 54, and so on. To get from the first to the second term, second to the third term, and third to the fourth term, we need to multiply by three, as two multiplied by three is six, six multiplied by three is 18, and 18 multiplied by three is 54.

This constant in the example three is known as the common ratio denoted by the letter π‘Ÿ. We let the first term of any geometric sequence be the letter π‘Ž or π‘Ž sub one. As we multiply this by the constant π‘Ÿ to get the next term, the second term, π‘Ž sub two, is equal to π‘Žπ‘Ÿ. The third term is equal to π‘Žπ‘Ÿ squared. This pattern continues such that the 𝑛th term is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. This gives us a general formula for the 𝑛th term of a geometric sequence. We will now look at how we can calculate the sum of a finite geometric sequence.

The sum of the first 𝑛-terms of a geometric sequence can be written as follows. 𝑆 sub 𝑛 is equal to π‘Ž plus π‘Žπ‘Ÿ plus π‘Žπ‘Ÿ squared, and so on, up to the last two terms of π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus two and π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. We will call this equation one. If we multiply each of the terms in this equation by π‘Ÿ, we get π‘Ÿ multiplied by 𝑆 sub 𝑛 is equal to π‘Žπ‘Ÿ plus π‘Žπ‘Ÿ squared plus π‘Žπ‘Ÿ cubed, and so on, such that the last two terms are π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one and π‘Ž multiplied by π‘Ÿ to the power of 𝑛. We will call this equation two and then subtract this equation from equation one.

On the left-hand side, we have 𝑆 sub 𝑛 minus π‘Ÿ multiplied by 𝑆 sub 𝑛. We can factor out 𝑆 sub 𝑛 such that this becomes 𝑆 sub 𝑛 multiplied by one minus π‘Ÿ. On the right-hand side, when subtracting, the π‘Žπ‘Ÿ terms will cancel, likewise, π‘Žπ‘Ÿ squared. In fact, all the terms will cancel apart from π‘Ž in equation one and π‘Ž multiplied by π‘Ÿ to the power of 𝑛 in equation two. This means that the right-hand side becomes π‘Ž minus π‘Ž multiplied by π‘Ÿ to the power of 𝑛. These two terms have a common factor of π‘Ž, so we can factor this out.

Next, we can divide both sides of our equation by one minus π‘Ÿ. This gives us 𝑆 sub 𝑛 is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 all divided by one minus π‘Ÿ. This formula enables us to calculate the sum of the first 𝑛-terms of a finite geometric sequence. We will now look at some questions where we need to apply this formula.

A geometric sequence has a first term of three and a common ratio of five. Find the sum of the first six terms.

We know that the sum of the first 𝑛-terms of a geometric sequence, written 𝑆 sub 𝑛, is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 all divided by one minus π‘Ÿ. In this question, we are told that the first term π‘Ž is equal to three. The common ratio π‘Ÿ is equal to five. And we are interested in the sum of the first six terms. Therefore, 𝑛 is equal to six. Substituting in these values, we have that 𝑆 sub six is equal to three multiplied by one minus five to the power of six all divided by one minus five. Five to the power of six or five to the sixth power is equal to 15625 and one minus five is equal to negative four. Typing this calculation into our calculator gives us 11718. The sum of the first six terms of the geometric sequence with first term three and common ratio five is 11718.

In our next question, we will need to calculate the common ratio π‘Ÿ and the number of terms 𝑛 before calculating the sum of the sequence.

Find the sum of the geometric sequence 16, negative 32, 64, and so on, all the way up to 256.

We know that the sum of any geometric sequence denoted 𝑆 sub 𝑛 is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 all divided by one minus π‘Ÿ. We can see immediately from the sequence that the value of the first term π‘Ž is 16. The second term is equal to π‘Ž multiplied by π‘Ÿ, and this is equal to negative 32. If we label these equations one and two, we can calculate the value of π‘Ÿ by dividing equation two by equation one. On the left-hand side, we have π‘Žπ‘Ÿ divided by π‘Ž, and on the right-hand side negative 32 divided by 16. As π‘Ž is not equal to zero, we can cancel this on the left-hand side. And negative 32 divided by 16 is negative two.

This value of π‘Ÿ makes sense as we multiply the first term 16 by negative two to get the second term negative 32. This also works to get from the second to third term. Negative 32 multiplied by negative two is 64. We know that the 𝑛th term of any geometric sequence written π‘Ž sub 𝑛 is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. To calculate the value of 𝑛, we can substitute our values of π‘Ž and π‘Ÿ and the 𝑛th term 256. This gives us the equation 256 is equal to 16 multiplied by negative two to the power of 𝑛 minus one. We can divide both sides by 16 such that 16 is equal to negative two to the power of 𝑛 minus one.

We know that negative two to the fourth power or negative two to the power of four is equal to 16. This means that 𝑛 minus one must be equal to four. Adding one to both sides of this equation gives us a value of 𝑛 equal to five. We now have values of π‘Ž, π‘Ÿ, and 𝑛. The sum of the first five terms is therefore equal to 16 multiplied by one minus negative two to the fifth power all divided by one minus negative two. This simplifies to 16 multiplied by one plus 32 all divided by three. Typing this into the calculator, we get an answer of 176. The sum of the geometric sequence 16, negative 32, 64, and so on, up to 256 is equal to 176.

In our next question, we need to calculate the number of terms in a geometric sequence.

The number of terms of a geometric sequence whose first term is 729, last term is one, and sum of all terms is 1093 is blank.

We’re told that the first term of our sequence π‘Ž sub one or π‘Ž is 729. The last term π‘Ž sub 𝑛 is equal to one. We know that π‘Ž sub 𝑛 is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. We are also told that the sum of all the terms 𝑆 sub 𝑛 is equal to 1093, where 𝑆 sub 𝑛 is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 all divided by one minus π‘Ÿ. Our aim in this question is to calculate the number of terms 𝑛.

Substituting in the value of π‘Ž, we see that 729 multiplied by π‘Ÿ to the power of 𝑛 minus one is equal to one. Dividing both sides of this equation by 729, π‘Ÿ to the power of 𝑛 minus one is equal to one over 729. Using one of our laws of exponents or indices, we can rewrite the left-hand side as π‘Ÿ to the power of 𝑛 over π‘Ÿ to the power of one, as π‘₯ to the power of 𝑝 minus π‘ž is equal to π‘₯ to the power of 𝑝 divided by π‘₯ to the power of π‘ž. We can then multiply both sides of this equation by π‘Ÿ, giving us π‘Ÿ sub 𝑛 is equal to one over 729 π‘Ÿ.

We will now clear some space and consider the second formula. This gives us 729 multiplied by one minus one over 729 π‘Ÿ all divided by one minus π‘Ÿ is equal to 1093. We can distribute the parentheses of the numerator on the left-hand side to give us 729 minus π‘Ÿ. Multiplying through by one minus π‘Ÿ gives us 729 minus π‘Ÿ is equal to 1093 multiplied by one minus π‘Ÿ. We can once again distribute the parentheses or expand the brackets giving us 1093 minus 1093π‘Ÿ. Subtracting 729 and adding 1093π‘Ÿ to both sides gives us 1092π‘Ÿ is equal to 364. We can then divide both sides by 1092 such that π‘Ÿ is equal to one-third. We can now substitute this back in to calculate the value of 𝑛.

One-third to the power of 𝑛 is equal to one over 729 multiplied by one-third. We know that three to the power of six is equal to 729. This means that one-third to the power of six is equal to one over 729. This means that we can rewrite the right-hand side of our equation as one-third to the power of six multiplied by one-third. Once again, using our laws of exponents, we can add the powers. Six plus one is equal to seven. As one-third to the power of 𝑛 is equal to one-third to the power of seven, then 𝑛 must be equal to seven. The number of terms of the geometric sequence whose first term is 729, last term is one, and sum of all terms is 1093 is seven.

In our final question, we will once again need to find the sum of 𝑛-terms of a geometric sequence.

Find the sum of the first seven terms of a geometric sequence given π‘Ž sub five is equal to negative eight multiplied by π‘Ž sub two and π‘Ž sub four plus π‘Ž sub six is equal to negative 64.

We know that the 𝑛th term of any geometric sequence denoted π‘Ž sub 𝑛 is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. The fifth term of the sequence will therefore be equal to π‘Ž multiplied by π‘Ÿ to the fourth power, and the second term π‘Ž sub two will be equal to π‘Žπ‘Ÿ. Likewise, the fourth term will be equal to π‘Žπ‘Ÿ cubed, and the sixth term π‘Ž sub six will be equal to π‘Ž multiplied by π‘Ÿ to the fifth power. We can therefore rewrite our two equations.

Firstly, we have π‘Ž multiplied by π‘Ÿ to the fourth power is equal to negative eight π‘Žπ‘Ÿ. As both π‘Ž and π‘Ÿ cannot be equal to zero, we can divide through by π‘Ž and π‘Ÿ. This leaves us with π‘Ÿ cubed is equal to negative eight. We can cube root both sides of this equation such that π‘Ÿ is equal to negative two. We can then substitute this value of π‘Ÿ into our second equation. π‘Žπ‘Ÿ cubed plus π‘Žπ‘Ÿ to the fifth power is equal to negative 64. Negative two cubed is equal to negative eight. So the first term becomes negative eight π‘Ž. Negative two to the fifth power is equal to negative 32. Our equation becomes negative eight π‘Ž plus negative 32π‘Ž is equal to negative 64.

The left-hand side simplifies to negative 40π‘Ž. We can then divide through by negative 40 such that π‘Ž is equal to eight over five or eight-fifths. This is also equal to the decimal 1.6. We now have a value of π‘Ÿ, a value of π‘Ž, and a value of 𝑛 equal to seven, as we need to calculate the sum of the first seven terms. In order to do this, we will use the formula 𝑆 sub 𝑛 is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 all divided by one minus π‘Ÿ. 𝑆 sub seven is therefore equal to 1.6 or eight over five multiplied by one minus negative two to the seventh power all divided by one minus negative two. Typing this into the calculator gives us an answer of 344 over five. As a decimal, this is equal to 68.8. The sum of the first seven terms of a geometric sequence given that π‘Ž sub five is equal to negative eight multiplied by π‘Ž sub two and π‘Ž sub four plus π‘Ž sub six is equal to negative 64 is 344 over five or 68.8.

We will now summarize the key points from this video. A geometric sequence has a first term π‘Ž and a common ratio π‘Ÿ. And each term is found by multiplying the previous term by this common ratio. The 𝑛th term of any geometric sequence written π‘Ž sub 𝑛 is equal to π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. And the sum of the first 𝑛-terms of a geometric sequence written 𝑆 sub 𝑛 is equal to π‘Ž multiplied by one minus π‘Ÿ to the 𝑛th power divided by one minus π‘Ÿ. We saw in this video that we can use these formulae to calculate the sum of the terms in a finite geometric sequence.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.