Question Video: Identifying Which Geometric Sequence Can Be Summed Up to Infinity Mathematics

Which of the following geometric sequences can be summed up to infinity? [A] 8 Γ— 6^(𝑛 βˆ’ 5) [B] 1/28, 5/28, 9/28, β‹…β‹…β‹… [C] 263, βˆ’789/5, 2367/25, β‹…β‹…β‹… [D] 510, βˆ’680, 2720/3, β‹…β‹…β‹… [E] βˆ’510, 680, βˆ’2720/3, β‹…β‹…β‹…

03:23

Video Transcript

Which of the following geometric sequences can be summed up to ∞?

And then, we’re given five sequences to choose from. We begin by recalling that a geometric sequence is one which has a common ratio between terms. The 𝑛th term of a geometric sequence is π‘Žπ‘Ÿ to the power of 𝑛 minus one, where π‘Ž is the first term and π‘Ÿ is the common ratio. Now, we can sum a geometric sequence to ∞ if it’s convergent. And this occurs when the modulus of π‘Ÿ or the absolute value of π‘Ÿ, where π‘Ÿ is the common ratio, is less than one.

Now, we’re told that all of these sequences are geometric sequences, and so our job is to find the common ratio. Let’s begin with sequence (A). That’s eight times six to the power of 𝑛 minus five. We’re going to need to write this in this form: π‘Ž times π‘Ÿ to the power of 𝑛 minus one. And we can use one of our rules for exponents to do so. We know that when we multiply two exponential terms which have the same base, we simply add their exponents. And so, we can reverse this and say that β€œWell, six to the power of 𝑛 minus five is the same as six to the power of negative four times six to the power of 𝑛 minus one.” Six to the power of negative four, though, is one over six to the power of four. And so, we can rewrite this as eight over six to the power of four times six to the power of 𝑛 minus one.

If we compare this to the general 𝑛th term of a geometric sequence, we see that π‘Ž, which is the first term in this sequence, is eight over six to the fourth power. π‘Ÿ, however, is six. Now, it’s quite clear to us that six is not less than one. We can, therefore, say that sequence (A) is not convergent, and so we cannot sum it up to ∞.

And so, we move on to sequence (B). This time, we recall that we can find the common ratio by dividing any term by the term that precedes it. And so, in this case, we can divide three twenty-eighths by one twenty-eighth. And since the denominators of these fractions is the same, that’s three divided by one, which is three. Once again, three is not less than one. And so, sequence (B) is not convergent and cannot be summed to ∞.

Let’s have a look at sequence (C) now. Once again, we’ll find the common ratio by dividing the second term by the first. Remember, we would get the same answer if we divided the third term by the second and so on. One method we have for dividing fractions is to create a common denominator. Now, if we multiplied 263 by five, we get 1315. So, 263 is equivalent to one thousand three hundred and fifteen fifths. And so, we then divide the numerators. And we find π‘Ÿ is equal to negative 789 over 1315. This value for π‘Ÿ is between negative one and one. And so, we can say that the absolute value of the common ratio in this sequence is less than one. This means it’s convergent and it can be summed to ∞.

Let’s just double-check the other two sequences. By dividing the second term by the first in sequence (D), we can see that the absolute value of the common ratio is not less than one, and so it can’t be (D). And in fact, we get the same value for the common ratio in sequence (E). And so that confirms to us that the only geometric sequence in this list that can be summed to ∞ is (C).

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