Question
If $a, b, c$ and $d$ are in G.P show that $\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)=(a b+b c+c d)^2$.

Answer

a, b, c, d are in G. P
$b=a r$
$\mathrm{c}=\mathrm{ar}^2$
$d=a r^3$
$\text { L. H. S }=\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)$
$=\left(a^2+a^2 r^2+a^2 r^4\right)\left(a^2 r^2+a^2 r^4+a^2 r^6\right)$
$=a^4 r^2\left(1+r^2+r^4\right)^2$
R. H.S. $=(a b+b c+c d)^2$
$=\left(a^2 r+a^2 r^3+a^2 r^5\right)^2$
$=a^4 r^2\left(1+r^2+r^4\right)^2$
H.p

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