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find 1st term a5= 405 a8= 10935

Guest Jul 10, 2017
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 #1
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+1

geometric sequence:

find 1st term a5= 405 a8= 10935

 

\(\begin{array}{|lcrl|} \hline a_1 = a &=& \ ?\\ a_5 = 405 &=& a\cdot r^4 \\ a_8 = 10935 &=& a\cdot r^7 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \frac{a_8}{a_5} = \frac{10935}{405} &=& \frac{a\cdot r^7 }{a\cdot r^4 } \\ \frac{10935}{405} &=& \frac{ r^7 }{ r^4 } \\ \frac{10935}{405} &=& r^3 \\ 27 &=& r^3 \\ \mathbf{3} & \mathbf{=} & \mathbf{r} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline a_5 = 405 &=& a\cdot r^4 \\ 405 &=& a\cdot 3^4 \\ 405 &=& a\cdot 81 \\ a &=& \frac{405}{81} \\ \mathbf{a} & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

 

The 1st term is 5.

 

laugh

heureka  Jul 10, 2017
 #2
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+1

a8=10,935

a5=405

a5=ar^4

a8=ar^7

 

ar^4 divided by ar^7 will equal 405/10,935

r^4-7 will equal .037

1/r^3 will equal .037

Therefore,  r=1/cube root of .037=1/.33    ~3

Therefore, a=405/r^4   =405/3^4    =405/81   =5

Guest Jul 10, 2017

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