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# question about geometric sequence problem

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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

Guest May 15, 2015

#1
+20680
+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.

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

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

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

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

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

heureka  May 15, 2015
#1
+20680
+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.

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

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

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

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

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

heureka  May 15, 2015