PLEASE!!!!! HELP ME!!!!!! Two terms in a geometric sequence are a5=15 and a6=1. What is the recursive rule that describes the sequence? A) a1=50,625; an=an−1⋅15 B) a1=11,390,625; an=an−1⋅15 C) a1=759,375; an=an−1⋅115 D) a1=225; an=an−1⋅5

1 answer
Question:

PLEASE!!!!! HELP ME!!!!!!
Two terms in a geometric sequence are a5=15 and a6=1.
What is the recursive rule that describes the sequence?
A) a1=50,625; an=an−1⋅15
B) a1=11,390,625; an=an−1⋅15
C) a1=759,375; an=an−1⋅115
D) a1=225; an=an−1⋅5

Answers

Answer:[tex] 15 = a_1 r^4 [/tex] (1)[tex] 1 = a_1 r^5[/tex] (2)If we divide equations (2) and (1) we got:[tex] \frac{r^5}{r^4}= \frac{1}{15}[/tex]And then [tex] r= \frac{1}{15}[/tex]And then we can find the value [tex] a_1[/tex] and we got from equation (1)[tex] a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375[/tex]And then the general term for the sequence would be given by:[tex] a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,... [/tex]And the best option would be:C) a1=759,375; an=an−1⋅(1/15)Step-by-step explanation:the general formula for a geometric sequence is given by:[tex] a_n = a_1 r^{n-1}[/tex]For this case we know that [tex] a_5 = 15, a_6 = 1[/tex]Then we have the following conditions:[tex] 15 = a_1 r^4 [/tex] (1)[tex] 1 = a_1 r^5[/tex] (2)If we divide equations (2) and (1) we got:[tex] \frac{r^5}{r^4}= \frac{1}{15}[/tex]And then [tex] r= \frac{1}{15}[/tex]And then we can find the value [tex] a_1[/tex] and we got from equation (1)[tex] a_1 = \frac{15}{r^4} = \frac{15}{(\frac{1}{15})^4} =759375[/tex]And then the general term for the sequence would be given by:[tex] a_n = 759375 (\frac{1}{15})^n-1 , n=1,2,3,4,... [/tex]And the best option would be:C) a1=759,375; an=an−1⋅(1/15)

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