Answer:
[tex]a_7=\frac{27}{4}[/tex]
Step-by-step explanation:
As the nth term of the Geometric profession is
[tex]a_{n}=a\,r^{n-1}[/tex]
Where a is the initial value or first term i.e. [tex]a=a_{1}[/tex] and common ratio r.
As the first term is
[tex]a_{1}=16[/tex]
and the 5th term will be
[tex]a_5=ar^{5-1}[/tex]
As
[tex]a_{5} =9[/tex]
So
[tex]9=ar^{5-1}[/tex]
[tex]9=ar^{4}[/tex]
[tex]9=(16)r^{4}[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}16[/tex]
[tex]\frac{16r^4}{16}=\frac{9}{16}[/tex]
[tex]r^4=\frac{9}{16}[/tex]
[tex]\mathrm{Rewrite\:the\:equation\:with\:}u=r^2\mathrm{\:and\:}u^2=r^4[/tex]
[tex]u^2=\frac{9}{16}[/tex]
[tex]\mathrm{Solve\:}\:u^2=\frac{9}{16}:\quad u=\sqrt{\frac{9}{16}},\:u=-\sqrt{\frac{9}{16}}[/tex]
[tex]u=\sqrt{\frac{9}{16}},\:u=-\sqrt{\frac{9}{16}}[/tex]
[tex]\mathrm{Substitute\:back}\:u=r^2,\:\mathrm{solve\:for}\:r[/tex]
[tex]\mathrm{Solve\:}\:r^2=\sqrt{\frac{9}{16}}:\quad r=\frac{\sqrt{3}}{2},\:r=-\frac{\sqrt{3}}{2}[/tex]
[tex]\mathrm{Solve\:}\:r^2=-\sqrt{\frac{9}{16}}:\quad r=i\frac{\sqrt{3}}{2},\:r=-i\frac{\sqrt{3}}{2}[/tex]
So, there are the following values of r
[tex]r=\frac{\sqrt{3}}{2},\:r=-\frac{\sqrt{3}}{2},\:r=i\frac{\sqrt{3}}{2},\:r=-i\frac{\sqrt{3}}{2}[/tex]
But, we will check only for the real values for r
For [tex]r=\frac{\sqrt{3}}{2}[/tex], the seventh term will be:
[tex]a_7=ar^{7-1}[/tex]
[tex]a_7=16\left(\frac{\sqrt{3}}{2}\right)^{7-1}[/tex]
[tex]a_7=16\left(\frac{\sqrt{3}}{2}\right)^{6}[/tex]
As
[tex]\left(\frac{\sqrt{3}}{2}\right)^{7-1}=\frac{3^3}{2^6}[/tex]
So
[tex]a_7=16\cdot \frac{3^3}{2^6}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}[/tex]
[tex]a_7=\frac{3^3\cdot \:16}{2^6}[/tex]
[tex]\mathrm{Factor}\:16:\quad 2^4[/tex]
[tex]a_{7} =\frac{2^4\cdot \:3^2}{2^4}[/tex]
[tex]\mathrm{Cancel\:}\frac{3^3\cdot \:2^4}{2^6}:\quad \frac{3^3}{2^2}[/tex]
[tex]a_7=\frac{3^3}{2^2}[/tex]
[tex]a_7=\frac{27}{2^2}[/tex]
[tex]a_7=\frac{27}{4}[/tex]
For [tex]r=-\frac{\sqrt{3}}{2}[/tex]
[tex]a_7=16\left(-\frac{\sqrt{3}}{2}\right)^{7-1}[/tex]
[tex]a_7=16\cdot \frac{3^3}{2^6}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}[/tex]
[tex]a_7=\frac{3^3\cdot \:16}{2^6}[/tex]
[tex]\mathrm{Factor}\:16:\quad 2^4[/tex]
[tex]a_{7} =\frac{2^4\cdot \:3^2}{2^4}[/tex]
[tex]\mathrm{Cancel\:}\frac{3^3\cdot \:2^4}{2^6}:\quad \frac{3^3}{2^2}[/tex]
[tex]a_7=\frac{3^3}{2^2}[/tex]
[tex]a_7=\frac{27}{2^2}[/tex]
[tex]a_7=\frac{27}{4}[/tex]
Therefore, for [tex]r=\frac{\sqrt{3}}{2}[/tex] and [tex]r=-\frac{\sqrt{3}}{2}[/tex],
[tex]a_7=\frac{27}{4}[/tex]
Keywords: geometric progression, geometric sequence, common ratio
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