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i need help this this question. in a geometric sequence the second term is 28 and the fifth term is 1792. find the 8th term.

step by step tutorial in how to solve this question would be appreciated very much :D 

 May 15, 2015

Best Answer 

 #1
avatar+26363 
+15

i need help this this question. in a geometric sequence the second term is 28 and the fifth term is 1792. find the 8th term.

$$Formula: \boxed{~ a_n=a\cdot r^{n-1} ~}$$

 

$$a_2=28=a\cdot{r}^1 \qquad a_5 = 1792=a\cdot r^4\qquad a_8 = a\cdot r^7$$

 

$$\dfrac{a_5}{a_2}=\dfrac{a\cdot r^4}{a\cdot r^1} = r^3\\\\\\
\small{\text{$
\begin{array}{rcl}
r^3 &=& \dfrac {1792}{28}\\\\
r &=& \sqrt[3]{\dfrac {1792}{28}}\\\\
r&=&\sqrt[3]{64}\\\\
r &=&4
\end{array}
$}}$$

 

$$a=\dfrac{28}{r}=\dfrac{28}{4}=7$$

 

$$\\a_8 = a\cdot r^7\\
a_8= 7\cdot 4^7\\
a_8 = 7\cdot 16384\\
a_8 = 114688$$

 May 15, 2015
 #1
avatar+26363 
+15
Best Answer

i need help this this question. in a geometric sequence the second term is 28 and the fifth term is 1792. find the 8th term.

$$Formula: \boxed{~ a_n=a\cdot r^{n-1} ~}$$

 

$$a_2=28=a\cdot{r}^1 \qquad a_5 = 1792=a\cdot r^4\qquad a_8 = a\cdot r^7$$

 

$$\dfrac{a_5}{a_2}=\dfrac{a\cdot r^4}{a\cdot r^1} = r^3\\\\\\
\small{\text{$
\begin{array}{rcl}
r^3 &=& \dfrac {1792}{28}\\\\
r &=& \sqrt[3]{\dfrac {1792}{28}}\\\\
r&=&\sqrt[3]{64}\\\\
r &=&4
\end{array}
$}}$$

 

$$a=\dfrac{28}{r}=\dfrac{28}{4}=7$$

 

$$\\a_8 = a\cdot r^7\\
a_8= 7\cdot 4^7\\
a_8 = 7\cdot 16384\\
a_8 = 114688$$

heureka May 15, 2015

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