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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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If the $4^{th},10^{th}\:\;and\:\:16^{th}$ terms of a G.P. are $x,y\:\;and\:\:z$ respectively. Prove that $x,y,z$ are in G.P.

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  • $n^{th}$ term of a G.P. =$t_n=a.r^{n-1}$
To prove $x,y,z$ are in G.P. we have to prove that $\large\frac{y}{x}=\frac{z}{y}$
or we should prove that $y^2=xz$
Given that in a G.P. $t_4=x,\:\:t_{10}=y,\:\:t_{16}=z$
$\Rightarrow\:a.r^{4-1}=a.r^3=x$
$a.r^{10-1}=a.r^9=y\:\:\:and\:\: a.r^{16-1}=a.r^{15}=z$
$\Rightarrow\:xz=(a.r^3).(a.r^{15})$
$\Rightarrow\:xz=a^2.r^{15+3}=a^2.r^{18}$......(i)
and
$y^2=(a.r^9)^2=a^2.r^{18}$.....(ii)
From (i) and (ii) $ y^2=xz$
$\therefore\:x,y,z$ are in G.P.
Hence Proved.
answered Feb 27, 2014 by rvidyagovindarajan_1
 

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