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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.

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  • $a^{mn}=(a^m)^n$
Given that $2$ terms are inserted between $3$ and $81$ to form a G.P.
Let the two numbers be $G_1$ and $G_2$.
$\Rightarrow\:3,G_1,G_2,81$ 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
$G_1G_2=3\times 81=3^5$.........(iii)
From (iii) $G_1=\large\frac{3^5}{G_2}$
Substituting this value of $G_1$ in (ii) we get
$G_2^2=81\times \large\frac{3^5}{G_2}$
$\Rightarrow\:G_2^3=81\times 3^5=3^9=(3^3)^3$
(Since $81=3^4)$
Taking cube root on both the sides we get
$\Rightarrow\:G_2=3^3=27$
Substituting the value of $G_2$ in (iii) we get
$G_1=\large\frac{3^5}{3^3}$$=3^2=9$
$\therefore\:$ The two numbers are $9$ and $27$
answered Mar 3, 2014 by rvidyagovindarajan_1
 

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