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find the 5th term of the sequence which has a first term of 17 and has a common ratio of 3.1.
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n=5-1 =4 17(3.1^4) =1570
\[a \times r^n\] where a is the first term and r is the common ratio n is the number of the term
A{1} = 17 ; and a geometric ratio o f 3.1 A{2} = 17 * 3.1 A{3} = 17 * (3.1 * 3.1) A{4} = 17 * (3.1 * 3.1 * 3.1); [3.1^3] A{n} = 17 * 3.1^(n-1)
A{5} = 17 * 3.1^(4)
I get 1569.98... if i did it right
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a(1) = 17 a(2)=52.7 a(3)=163.37 a(4)=506.447 a(4)=1569.986
except the last index should be 5! ha ha ha
lol
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