Browse Questions

# If $a,b,c,d$ are in G.P, then show that $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

Toolbox:
• The first four terms of a G.P. are assumed as $a,ar,ar^2,ar^3$
• $(ab)^m=a^m.b^m$
• $(a^m)^n=a^{mn}$
Step 1
The first four terms of a G.P. are assumed as $a,ar,ar^2,ar^3$
Given that $a,b,c,d$ are in G.P.
$\Rightarrow\:$ Let $a=a,\:\:ar=b,\:\:ar^2=c\:\:and\:\:ar^3=d$
$\Rightarrow\:b^2=(a.r)^2=a^2.r^2$
$c^2=(a.r^2)^2=a^2.r^4$ and
$d^2=(a.r^3)^2=a^2.r^6$
$\Rightarrow\:a^2+b^2+c^2=a^2+a^2.r^2+a^2.r^4=a^2(1+r^2+r^4)$
and
Step 2
$b^2+c^2+d^2=a^2.r^2+a^2.r^4+a^2r^6=a^2.r^2(1+r^2+r^4)$
Step 3
L.H.S.= $(a^2+b^2+c^2)(b^2+c^2+d^2)=a^2(1+r^2+r^4)\times a^2.r^2(1+r^2+r^4)$
$\Rightarrow\:$ L.H.S.$=a^4.r^2.(1+r^2+r^4)^2$.........(i)
Step 4
$ab=a(a.r)=a^2r$
$bc=(ar)(ar^2)=a^2.r^3$
$cd=(ar^2)(ar^3)=a^2.r^5$
$\Rightarrow\:(ab+bc+cd)=a^2.r+a^2.r^3+a^2.r^5$
$=a^2.r(1+r^2+r^4)$
$\therefore\:(ab+bc+cd)^2=(a^2.r(1+r^2+r^4))^2$
$=a^4.r^2(1+r^2+r^4)^2$
$\Rightarrow\:$ R.H.S.=$(ab+bc+cd)^2=a^4.r^2(1+r^2+r^4)^2$.........(ii)
From (i) and (ii) L.H.S.=R.H.S.
Hence proved.