In Exercises 63–64, find a2 and a3 for each geometric sequence.

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Question

In Exercises 63–64, find a2 and a3 for each geometric sequence.

8, a2, a3, 27

Accepted Answer

Hey, everyone here, we are asked to determine the missing terms in the given geometric sequence. Now, before we can begin solving this problem, we first need to recall the formula for the general term of a geometric sequence, which is a sub N is equal to a sub one times R to the power of N minus one where a sub one is the first term. So we can go ahead and identify our ace of one in reference to our given sequence. So since we know this is the first term, we can look at our first term in the sequence which is 64. So our ace of one is simply equal to 64. And lastly, we need to use this last piece of given information so that we can eventually solve four R. So this last piece of information we have the value of 125. Now we need to note that this term here is the fourth term in the sequence. And from this, we know that a sub four is equal to 125. And now with this information identified, we can go ahead and plug it into our formula and solve four R. So instead of a sub N, we'll have a sub four on the left hand side. So plugging in that value, we have, we have 125 is equal to, then we have a sub one which we identified as 64 times R. And in the form formula R is to the power of N minus one. But our subscript here, which is on the left hand side of the formula is the value for from our fourth term. So R is going to be to the power of four minus one since N is replaced by the value of the subscript. So with this, we can just simplify and solve for R. So first, we can divide both sides by 64 giving us R cubed is equal to 125 five divided by 64. Now we take the third route from both sides and we see that R is equal to five fourths. And now with all of this information, we can go ahead and find our missing values. So starting with a sub two, a sub two and using the formula is equal to a sub one which again is 64 times R which is five fourths as we found to the power of N minus one. And in this case, our subscript is two. So R is to the power of two minus one. Now we can just simplify and we see that a sub two is equal to 80. Now, we can find our last term here, which is a sub three. A sub three is equal to a sub one, which is again 64 times R which is five fourths to the power of three. In this case, three is N minus one. And now simplifying, we see that a sub three is equal to 100. And with this, we have identified our two missing terms and we are left with answer choice. C Thanks for watching. I hope you found this video helpful and I'll see you again next time.

In Exercises 63–64, find a2 and a3 for each geometric sequence. 8, a2, a3, 27

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