Show that the sequence - defined by an-23 n - n - N is a G-P-

We #160;have,an=23n, #160;n #8712;NPutting #160;n=1, #160;2, #160;3, #160;...a1=231=23, #160;a2=232=29, #160;a3=233=227 #160;and #160;so ...

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Standard IV

Mathematics

Language Patterns

Show that the...

Question

Show that the sequence <a_n>, defined by a_n = $\frac{2}{3^{n}}$ , n ϵ N is a G.P.

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Solution

$We have, a_{n} = \frac{2}{3^{n}}, n \in N Putting n = 1, 2, 3, . . . a_{1} = \frac{2}{3^{1}} = \frac{2}{3}, a_{2} = \frac{2}{3^{2}} = \frac{2}{9}, a_{3} = \frac{2}{3^{3}} = \frac{2}{27} and so on . Now, \frac{a_{2}}{a_{1}} = \frac{\frac{2}{9}}{\frac{2}{3}} = \frac{1}{3}, \frac{a_{3}}{a_{2}} = \frac{\frac{2}{27}}{\frac{2}{9}} = \frac{1}{3} and so on . ∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = . . . = \frac{1}{3} So, the sequence is an G . P ., where \frac{2}{3} is the first term and \frac{1}{3} is the common ratio .$

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Show that the sequence <an>, defined by an = 23n, n ϵ N is a G.P.

We have,an=23n, n∈NPutting n=1, 2, 3, ...a1=231=23, a2=232=29, a3=233=227 and so on.Now, a2a1=2923=13, a3a2=22729=13 and so on.∴a2a1=a3a2= ... =13So, the sequence is an G.P., where 23 is the first term and 13 is the common ratio.

Show that the sequence <a_n>, defined by a_n = $\frac{2}{3^{n}}$ , n ϵ N is a G.P.