Miscellaneous Examples (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths

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Chapter 8 - Sequences & Series | NCERT Solutions for Class 11 Maths

Miscellaneous Example 1

If $a, b, c, d$ and $p$ are different real numbers such that $\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0$, then show that $a, b, c$ and $d$ are in G.P.

Solution

Given 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) \leq 0
$

But L.H.S.
$
=\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),
$
which gives $(a p-b)^2+(b p-c)^2+(c p-d)^2 \geq 0$
Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), we have, $\quad(a p-b)^2+(b p-c)^2+(c p-d)^2=0$
or
$
a p-b=0, b p-c=0, c p-d=0
$

This implies that $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
Hence $a, b, c$ and $d$ are in G.P.