Solve this following


Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c$ $+\mathrm{cd}) \mathrm{p}+\left(\mathrm{b}^{2}+\mathrm{c}^{2}+\mathrm{d}^{2}\right)=0 .$ Then :




  1. a,c,p are in G.P.

  2. $\mathrm{a}, \mathrm{c}, \mathrm{p}$ are in A.P.

  3. $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in G.P.

  4. $a, b, c, d$ are in A.P.

Correct Option: , 3


$\left(a^{2}+b^{2}+c^{2}\right) p^{2}+2(a b+b c+c d) p+b^{2}+c^{2}+d^{2}$


$\Rightarrow\left(\mathrm{a}^{2} \mathrm{p}^{2}+2 \mathrm{abp}+\mathrm{b}^{2}\right)+\left(\mathrm{b}^{2} \mathrm{p}^{2}+2 \mathrm{bcp}+\mathrm{c}^{2}\right)+$

$\left(\mathrm{c}^{2} \mathrm{p}^{2}+2 \mathrm{cdp}+\mathrm{d}^{2}\right)=0$

$\Rightarrow(a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$

This is possible only when

$a p+b=0$ and $b p+c=0$ and $c p+d=0$


or $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

$\therefore a, b, c, d$ are in G.P.


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