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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Insert $2$ numbers between $3$ and $81$ so that the resulting sequence is a G.P. Find the two numbers.

$\begin{array}{1 1}9\;and\;27 \\ 6\;and\;27 \\ 6\;and \;36 \\ 9\;and\;36 \end{array} $

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  • $a^{mn}=(a^m)^n$
Given that $2$ numbers are inserted between $3$ and $81$
so that the resulting sequence is a G.P.
Let the $2$ numbers be $G_1,G_2$
$\Rightarrow\:3,G_1,G_2,81$ are in G.P.
$\Rightarrow\:$ common ratio $r=\large\frac{G_1}{3}=\frac{G_2}{G_1}=\frac{81}{G_2}$
$\Rightarrow\:G_1^2=3G_2$,.....(i)
$G_2^2=81G_1$......(ii) and
$G_1.G_2=3\times 81=3\times 3^4=3^5$.....(iii)
From (i) $G_2=\frac{G_1^2}{3}$
Substituting this value of $G_2$ in (iii) we get
$G_1.\large\frac{G_1^2}{3}$$=3^5$
$\Rightarrow\:G_1^3=3\times 3^5=3^6$
Taking cube root on both the sides
$G_1=(3^6)^{1/3}=3^2=9$
Substituting the value of $G_1$ in (i) we get
$9^2=3.G_2$
$\Rightarrow\:G_2=\large\frac{81}{3}$$=27$
$\therefore $ the two numbers inserted are $9$ and $27$.
answered Mar 4, 2014 by rvidyagovindarajan_1
 

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