Show that the sequence - defined by a n -2-3 n - n - N is a G-P-

We have, a_n = 2/3^n, n ϵ N Putting n = 1, 2, 3, .... a_1 = 2/3^1 = 2/3, a_2 = 2/3^2 = 2/9, a^3 = 2/3^3 = 2/27 and so on. Now, a_2/a_1 = 2/9/2/3 = 1/3, a_3/a_2 ...

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Standard XII

Mathematics

Geometric Progression

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,

$a_{n} = \frac{2}{3^{n}}, n ϵ N$

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, nϵN is a G.P.

We have, an=23n,nϵN Putting 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=....=13 So, 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.