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# find the general term of a geometric sequence

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27=u1r^3/192=u1r^6

Guest Jun 22, 2017

#1
+90988
+2

27=u1r^3/192=u1r^6

$$27=\frac{u_1r^3}{192}=u_1r^6\\ \frac{r^3}{192}=r^6\\ \frac{1}{192}=r^3\\ r=\frac{1}{\sqrt[3]{192}}\\ r=\frac{1}{4\sqrt[3]{3}}\\ 27=u_1 r^6\\ 27=\frac{u_1}{ 192^2}\\ u_1=27*192^2\\ u_1=995328$$

$$T_n=\dfrac{995328}{(\sqrt[3]{192})^{n-1}}\\~\\ \boxed{T_n=\dfrac{995328}{192^{((n-1)/3)}}}\\~\\ check\\ \frac{u_1r^3}{192}=\frac{995328*(192^{-1/3})^3}{192}=\frac{995328}{192^2}=27\\ u_1r^6=995328*((192)^{-1/3})^6\\ u_1r^6=995328*(192)^{-2}\\ u_1r^6=27\\ excellent$$

Melody  Jun 22, 2017
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#1
+90988
+2

27=u1r^3/192=u1r^6

$$27=\frac{u_1r^3}{192}=u_1r^6\\ \frac{r^3}{192}=r^6\\ \frac{1}{192}=r^3\\ r=\frac{1}{\sqrt[3]{192}}\\ r=\frac{1}{4\sqrt[3]{3}}\\ 27=u_1 r^6\\ 27=\frac{u_1}{ 192^2}\\ u_1=27*192^2\\ u_1=995328$$

$$T_n=\dfrac{995328}{(\sqrt[3]{192})^{n-1}}\\~\\ \boxed{T_n=\dfrac{995328}{192^{((n-1)/3)}}}\\~\\ check\\ \frac{u_1r^3}{192}=\frac{995328*(192^{-1/3})^3}{192}=\frac{995328}{192^2}=27\\ u_1r^6=995328*((192)^{-1/3})^6\\ u_1r^6=995328*(192)^{-2}\\ u_1r^6=27\\ excellent$$

Melody  Jun 22, 2017
#2
+78575
0

Very nice, Melody.....!!!!

CPhill  Jun 22, 2017

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