Sequences and Series 3 Question 11
11. If $a, b, c, d$ and $p$ are distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p$
$$ +\left(b^{2}+c^{2}+d^{2}\right) \leq 0, \text { then } a, b, c, d $$
(a) are in $AP$
(b) are in GP
(c) are in HP
(d) satisfy $a b=c d$
(1987, 2M)
Show Answer
Answer:
Correct Answer: 11. (b)
Solution:
- Here, $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p$
$$ \begin{gathered} \Rightarrow\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right) \\ \quad+\left(c^{2} p^{2}-2 c d p+d^{2}\right) \leq 0 \\ \Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2} \leq 0 \end{gathered} $$
$$ +\left(b^{2}+c^{2}+d^{2}\right) \leq 0 $$
[since, sum of squares is never less than zero] Since, each of the squares is zero.
$$ \begin{array}{ll} \therefore & (a p-b)^{2}=(b p-c)^{2}=(c p-d)^{2}=0 \\ \Rightarrow & p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c} \end{array} $$
$\therefore \quad a, b, c, d$ are in GP.
