Video: Finding the Infinite Geometric Series given the Sum of All Terms and the Sum of Certain Terms

Find the possible infinite geometric series where the sum is βˆ’21 and the sum of the first and second terms is βˆ’35/3.

06:01

Video Transcript

Find the possible infinite geometric series where the sum is negative 21 and the sum of the first and second terms is negative 35 over three.

Let’s begin by recalling what we know about a geometric series. The first term of any geometric series is denoted by the letter π‘Ž. The common ratio of a geometric series is equal to π‘Ÿ. This means that the terms of a geometric series are π‘Ž, π‘Žπ‘Ÿ, π‘Žπ‘Ÿ squared, and so on. The sum of the first 𝑛 terms of any geometric series is equal to π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 divided by one minus π‘Ÿ. The sum to ∞ of a geometric series is equal to π‘Ž divided by one minus π‘Ÿ. We can only calculate the sum to ∞ of a geometric series when the modulus or absolute value of π‘Ÿ is less than one.

We’re told in this question that the sum to ∞ is equal to negative 21. This means that π‘Ž divided by one minus π‘Ÿ equals negative 21. Multiplying both sides of this equation by one minus π‘Ÿ gives us π‘Ž is equal to negative 21 multiplied by one minus π‘Ÿ. We will call this equation one. We’re also told that the sum of the first and second terms is equal to negative 35 over three. This is the same as the sum of the first two terms. Therefore, 𝑠 of two is equal to negative 35 over three. π‘Ž multiplied by one minus π‘Ÿ to the power of 𝑛 divided by one minus π‘Ÿ is equal to negative 35 over three.

To simplify this equation, we can cross multiply. We can multiply both sides by one minus π‘Ÿ and both sides by three. As 𝑛 is equal to two, this gives us three π‘Ž multiplied by one minus π‘Ÿ squared is equal to negative 35 multiplied by one minus π‘Ÿ. We will call this equation two. We can now solve these two simultaneous equations to calculate our values for π‘Ÿ and π‘Ž. Let’s begin by dividing equation two by equation one.

On the left-hand side, the π‘Žβ€™s cancel. On the right-hand side, the one minus π‘Ÿβ€™s cancel. We’re, therefore, left with three multiplied by one minus π‘Ÿ squared is equal to 35 over 21. We will now clear some space to solve this equation. At this point, we could distribute the parentheses by multiplying three by one and three by negative π‘Ÿ squared. In this case, however, it is easier to divide both sides by three. The left-hand side becomes one minus π‘Ÿ squared. The right-hand side becomes 35 over 63.

Rearranging this equation, we get π‘Ÿ squared is equal to one minus 35 over 63. π‘Ÿ squared is, therefore, equal to 28 over 63. Both of these numbers are divisible by seven. 28 divided by seven is four and 63 divided by seven is equal to nine. Finally, square rooting both sides of this equation gives us π‘Ÿ is equal to the positive or negative square root of four over nine. When square rooting a fraction, we can square root the numerator and denominator separately. The square root of four is two and the square root of nine is three. Therefore, π‘Ÿ is equal to the positive or negative two-thirds.

We can now substitute each of these values into equation one to calculate π‘Ž. When π‘Ÿ is equal to positive two-thirds, π‘Ž is equal to negative 21 multiplied by one minus two-thirds. One minus two-thirds is equal to one-third. And one-third of negative 21 is negative seven. When π‘Ÿ is equal to negative two-thirds, π‘Ž is equal to negative 21 multiplied by one minus negative two-thirds. Subtracting a negative number is the same as adding that number. So, we have negative 21 multiplied by one plus two-thirds. One plus two-thirds is equal to five-thirds and five-thirds multiplied by negative 21 is negative 35.

We now have two possible series: a first term of negative seven and a common ratio of two-thirds or a first term of negative 35 and a common ratio of negative two-thirds. The first option gives us the series negative seven, negative 14 over three, negative 28 over nine, and so on. The second option gives us the series negative 35, 70 over three, negative 140 over nine, and so on. These are the two possible infinite geometric series in this question.

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