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What is the 9th term of the geometric sequence where a1 = 625 and a3 = 25?
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First get your formula. ab^n-1 You have the first term, so you can solve a. Then, when you have the a, you can use the third term to solve b. Then just put 9 into the formula (in n's place)
I am not getting it... so I use a1 as a and a3 as b?
Nope. The formula: \[T _{n}^{}=ab^{n-1}\] Now, they give you the first term. so: \[T _{1}^{} = ab^{1-1} = 625\] Now it's possible to get the a value, since b^0 = 1 a = 625. Now, you need to get b. Follow the same steps as I did, but use the third term now. and use the value you just got for a, which is 625
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