Browse Questions

# The $5^{th},8^{th}\:and\:11^{th}$ terms of a $G.P.$ are $p,q\:and\:s$ respectively. Show that $q^2=ps$

Toolbox:
• $n^{th}$ term of a $G.P.=t_n=a.r^{n-1}$ where $a=1^{st} term,\:\:r=common\:ratio.$
Given that $5^{th},8^{th}\:and\:11^{th}$ terms of a G.P. are $p,q\:and\:s$
$i.e., t_5=p,\:\:t_8=q\:\:and\:\:t_{11}=s$
$\Rightarrow\:a.r^{5-1}=a.r^4=p$,
$a.r^{8-1}=ar^7=q\:\:and\:\:a.r^{11-1}=a.r^{10}=s$
$\therefore \:q^2=\big[a.r^7\big]^2=a^2.r^{14}$ and
$ps=[a.r^4].[a.r^{10}]=a^2.r^{4+10}=a^2.r^{14}$
$\Rightarrow\:q^2=ps$
Hence proved.