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If $a, b, c$ and $d$ are in G.P. show that
$\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)=(a b+b c+c d)^2$
$\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)=(a b+b c+c d)^2$
Solution:
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Verified Answer
Let $r$ be the common ratio of the G.P $a, b, c, d$.
Then $b=a r, c=a r^2$ and $d=a r^3$
$\begin{aligned} \mathrm{LHS} &=\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right) \\ &=\left(a^2+a^2 r^2+a^2 r^4\right)\left(a^2 r^2+a^2 r^4+a^2 r^6\right) \\ &=a^4 r^2\left(1+r^2+r^4\right)^2 \\ \mathrm{RHS} &=(a b+b c+c d)^2=\left(a^2 r+a^2 r^3+a^2 r^5\right) \\ &=a^4 r^2\left(1+r^2+r^4\right)^2 \\ \text { Hence } &\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right) \\ &=(a b+b c+c d)^2 \end{aligned}$
Then $b=a r, c=a r^2$ and $d=a r^3$
$\begin{aligned} \mathrm{LHS} &=\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right) \\ &=\left(a^2+a^2 r^2+a^2 r^4\right)\left(a^2 r^2+a^2 r^4+a^2 r^6\right) \\ &=a^4 r^2\left(1+r^2+r^4\right)^2 \\ \mathrm{RHS} &=(a b+b c+c d)^2=\left(a^2 r+a^2 r^3+a^2 r^5\right) \\ &=a^4 r^2\left(1+r^2+r^4\right)^2 \\ \text { Hence } &\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right) \\ &=(a b+b c+c d)^2 \end{aligned}$
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