Let a, b, c, d and p be any non zero


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

  1. (1) $a, c, p$ are in A.P.

  2. (2) $a, c, p$ are in G.P.

  3. (3) $a, b, c, d$ are in G.P.

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

Correct Option: , 3


Rearrange given equation, we get

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

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

$\therefore a p-b=b p-c=c p-d=0$

$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ $\therefore a, b, c, d$ are in G.P.

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