Video: Finding the Terms of a Geometric Sequence given Its General Term

Find the first five terms of the sequence whose 𝑛th term is given by π‘Ž_(𝑛) = cos ((11𝑛)/6 πœ‹), where 𝑛 β‰₯ 1.

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Video Transcript

Find the first five terms of the sequence whose 𝑛th term is given by π‘Ž sub 𝑛 equals cos 11𝑛 over six πœ‹, where 𝑛 is greater than or equal to one.

In order to calculate the first five terms of any sequence starting with 𝑛 equals one, we substitute the values one, two, three, four, and five into our expression. π‘Ž sub one is therefore equal to cos of 11 multiplied by one over six πœ‹. π‘Ž sub two is equal to cos of 11 multiplied by two divided by six πœ‹. These simplify to cos of 11 over six πœ‹ and cos of 22 over six πœ‹, respectively. π‘Ž three is equal to cos of 33 over six πœ‹ as 11 multiplied by three is 33. Likewise, π‘Ž sub four is equal to cos of 44 over six πœ‹, and π‘Ž sub five is equal to cos 55 over six πœ‹.

At this stage, we could proceed in a few different ways. We could just type all five expressions into the calculator, ensuring that we were in radian mode. This would give us the first five terms of the sequence. Alternatively, we could cancel the fractions where possible; 22 over six, 33 over six, and 44 over six can all be canceled. Instead of doing either of these things, we will look at the properties of the cosine graph and some of our special angles. This will allow us to calculate the answers without using a calculator.

We recall that cos of 30 degrees is equal to root three over two. As 180 degrees is equal to πœ‹ radians, then cos of πœ‹ by six radians is also equal to root three over two. cos of πœ‹ over three or 60 degrees is equal to one-half. As πœ‹ over three is equivalent to two πœ‹ over six, cos of two πœ‹ over six equals one-half. We also know that cos of πœ‹ over two or 90 degrees is equal to zero. This means that cos of three πœ‹ over six equals zero. cos of four πœ‹ over six is equal to negative one-half, and cos of five πœ‹ by six is negative root three over two. As the cosine graph is periodic with a period of two πœ‹, we can calculate the value of cos 11 over six πœ‹, cos 22 over six πœ‹, and so on using these values and the pattern.

By considering our CAST diagram, we can find the cosine of all multiples of πœ‹ over six that are equal to root three over two. Two πœ‹ minus πœ‹ over six is equal to 11πœ‹ over six. As already mentioned, as the cosine graph is periodic and has a period of two πœ‹, we can add two πœ‹ to both of these values to find an infinite number of solutions. cos of 13πœ‹ over six and cos of 23πœ‹ over six are both equal to root three over two. As the first value π‘Ž sub one in our sequence was cos 11 over six πœ‹, this is equal to root three over two. Repeating this process, we see that cos of two over six πœ‹ and cos of 10 over six πœ‹ are equal to one-half. Once again, we can add two πœ‹ or 12 over six πœ‹ to both of these, giving us 14 over six πœ‹ and 22 over six πœ‹.

π‘Ž sub two, cos of 22 over six πœ‹, is equal to one-half. We could continue this method to calculate cos of 33 over six πœ‹ which equals zero, cos of 44 over six πœ‹ which is negative one-half, and cos of 55 over six πœ‹ which is equal to negative root three over two. As previously mentioned, there is no problem with typing the five expressions into our calculator in radian mode to get the first five terms root three over two, one-half, zero, negative one-half, and negative root three over two. It is important, however, to recall our special angles as these will save us time when dealing with some problems.

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