Show that the products of the corresponding terms of the sequences a, ar, ar^2,............ a r^ n-1 and A, A R, A R^2............. A R^ n-1 form a G.P. and find the common ratio.

Multiplying the corresponding terms of the given sequences, we have( a A ) , ( a r AR ) _ : ( a r ^ 2 AR ^ 2 ),........, ( a r ^ n - 1 AR ^ n - 1 ) (A), (a A r^R ), (a A r^2 R^2 ), , (a A r^ n-1 R^ n-1 ) are in G.P. Now a_2 a_1 = aArR aA = rR and a_3 a_2 = aA r^2 R^2 aArR = rR Since the ratio of the two succeeding terms are same, the resulting sequence is also in G.P and common ratio = a A r R a A = r R

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Show that the products of the corresponding terms of the sequences $\mathrm{a}, \mathrm{ar}, \mathrm{ar}^2$,............ $a r^{n-1}$ and $A, A R, A R^2$............. $A R^{n-1}$ form a G.P. and find the common ratio.

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Answer

Multiplying the corresponding terms of the given sequences, we have$( a \times \mathrm { A } ) , ( a r \times \mathrm { AR } ) _ { : } \left( a r ^ { 2 } \times \mathrm { AR } ^ { 2 } \right),$........,$\left( a r ^ { n - 1 } \times \mathrm { AR } ^ { n - 1 } \right)$
$\Rightarrow(A),\left(a A r^R\right),\left(a A r^2 R^2\right), \ldots \ldots,\left(a A r^{n-1} R^{n-1}\right)$ are in G.P.
Now ${{{a_2}} \over {{a_1}}} = {{aArR} \over {aA}} = rR$ and ${{{a_3}} \over {{a_2}}} = {{aA{r^2}{R^2}} \over {aArR}} = rR$
Since the ratio of the two succeeding terms are same, the resulting sequence is also in G.P
and common ratio = $\frac { a \mathrm { A } r \mathrm { R } } { a \mathrm { A } } = r \mathrm { R }$

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