$\frac{a_{n+1}}{a_{n}}$ , for all n $\epsilon$ N ------(i)

Let 'k' be any non-zero constant by which we are multiplying each term of G.P.so we get,

$ka_{1}, ka_{2},ka_{3},...,ka_{n}$

So equation (i)⇒ $\frac{ka_{n+1}}{ka_{n}}$ , for all n $\epsilon$ N

which is same as $\frac{a_{n+1}}{a_{n}}$

&ther4; the new sequence formed is also in G.P. with common ratio 'r'.

$\frac{a_{n+1}}{a_{n}}$ , for all n $\epsilon$ N ------(i)

Now the sequence formed by reciprocal will be

$\frac {1}{a_{1}}, \frac {1}{a_{2}},\frac {1}{a_{3}},...,\frac {1}{a_{n}}$

The common ratio = $\frac{\frac{1}{a_{n+1}}}{\frac{1}{a_{n}}} = \frac{a_{n+1}}{a_{n}}=\frac{1}{r}$

So the new sequence is in G.P. with common ratio $\frac{1}{r}$

$\frac{a_{n+1}}{a_{n}}$ , for all n $\epsilon$ N ------(i)

Let each term of G.P. is raised by exponent 'k' then the sequence becomes

$a_{1}^{k}, a_{2}^{k},a_{3}^{k},...,a_{n}^{k}$

$\frac{a_{n+1}^{k}}{a_{n}^{k}} = \left ( \frac{a_{n+1}}{a_{n}} \right )^{k} = r^{k}$ for all n $\epsilon$ N

∴ $a_{1}^{k}, a_{2}^{k},a_{3}^{k},...,a_{n}^{k}$ is in G.P with common ration $r^{k}$

kth term from the beginning = $a_{k} = a_{1}r^{k - 1}$

kth term from the end = $a_{n - k + 1} = a_{1}r^{n - k}$

∴ product of kth term from the beginning and kth term from the end

= $a_{1}r^{k - 1} \times a_{1}r^{n - k}$

= $a_{1}^{2} r^{k - 1 + n - k} $

= $a_{1}^{2} r^{n -1} $ for k = 2,3,4,...(n -1)

Hence the product of the terms equidistant from the beginning and the end is always same and is equal to the product of first and the last term.

∴ common ration = $\frac{b}{a} = \frac{c}{a} $

∴ $b^{2}$ = ac

From properties of geometric progression to Home

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