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

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Accepted Answer

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> [!TLDR]
> Using the G.P. relations $b^2=ac$, $bd=c^2$ and $ad=bc$, the identity $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$ is proved.
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> Since $a,b,c,d$ are in G.P. with common ratio $r$: $b=ar,\ c=ar^2,\ d=ar^3$. From these:
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> $b^2=a^2r^2=a(ar^2)=ac$
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> $bd=(ar)(ar^3)=a^2r^4=(ar^2)^2=c^2$
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> $ad=a(ar^3)=a^2r^3=(ar)(ar^2)=bc$
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> Now use the algebraic identity (Lagrange's identity applied to $(a,b,c)$ and $(b,c,d)$):
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> $(a^2+b^2+c^2)(b^2+c^2+d^2)-(ab+bc+cd)^2=(ac-b^2)^2+(ad-bc)^2+(bd-c^2)^2$
> …

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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

Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Precise Definition

The $n$ -th Term

Sum of $n$ Terms

Sum of an Infinite GP

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Q.If a,b,c,da, b, c, da,b,c,d are in G.P., then show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2(a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2

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

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