Video Transcript
Find the geometric sequence given the sum of all the terms is 3339, the last term is 1696, and the common ratio is two.
In order to answer this question, we will need to recall two formulae. First, 𝑎 sub 𝑛 or the 𝑛th term is equal to 𝑎 multiplied by 𝑟 to the power of 𝑛 minus one, where 𝑎 is the first term and 𝑟 is the common ratio of the geometric sequence. The second formula tells us that the sum of the first 𝑛 terms, written 𝑠 sub 𝑛, is equal to 𝑎 multiplied by one minus 𝑟 to the power of 𝑛 or divided by one minus 𝑟. When the common ratio is greater than one, as in this case, we tend to rewrite this as 𝑎 multiplied by 𝑟 to the power of 𝑛 minus one all divided by 𝑟 minus one.
We are told that the last term is equal to 1696. Using the fact that 𝑟 is equal to two, this gives us 𝑎 multiplied by two to the power of 𝑛 minus one is equal to 1696. Our laws of indices or exponents mean that we can rewrite two to the power of 𝑛 minus one as two to the power of 𝑛 divided by two to the power of one. We can then multiply both sides of this equation by two such that 𝑎 multiplied by two to the power of 𝑛 is equal to 3392. As we have two unknowns in this equation, we will stop here and call this equation one.
We are also told in the question that the sum of all the terms is 3339. Distributing the parentheses or expanding the brackets of our formula along with substituting 𝑟 is equal to two gives us 𝑎 multiplied by two to the power of 𝑛 minus 𝑎 all divided by two minus one is equal to 3339. As two minus one is equal to one, we have 𝑎 multiplied by two to the power of 𝑛 minus 𝑎 is equal to 3339. We will call this equation two.
We can now substitute our value of 𝑎 multiplied by two to the power of 𝑛 into equation two. 3392 minus 𝑎 is equal to 3339. We can add 𝑎 and subtract 3339 from both sides of this equation. This gives us a value of 𝑎 equal to 53. The first term of the geometric sequence is 53. The geometric sequence contains the terms 53, 106, 212, and so on all the way up to 1696.
