Question

If $$a,b,c,d$$ and $$p$$ are real numbers such that $$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0$$, then, $$a,b,c$$ and $$d$$
A. are in A.P.
B. are in G.P.
C. are in H.P.
D. satisfy $$ab=cd$$

Answer 1

Hint: First, the left side of the given inequality is arranged to the form of the sum of the squares and then it is examined whether $$a,b,c,d$$ are in A.P., G.P., H.P. or $$ab=cd$$.

Formula Used:
If $$a,b,c,d$$ are in A.P., then $$b-a=c-b=d-c$$.
If $$a,b,c,d$$ are in G.P., then $${\displaystyle \frac{b}{a}}={\displaystyle \frac{c}{b}}={\displaystyle \frac{d}{c}}$$.
If $$a,b,c,d$$ are in H.P., then $${\displaystyle \frac{1}{a}},{\displaystyle \frac{1}{b}},{\displaystyle \frac{1}{c}},{\displaystyle \frac{1}{d}}$$ are in A.P.

Complete step by step solution:
 We have been given that $$a,b,c,d$$ and $$p$$ are real numbers such that $$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0$$
Rearrange the left side of the given in-equation in the form of sum of the squares
$$\begin{array}{l}(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0\\ \Rightarrow a^2{p}^2+b^2{p}^2+c^2{p}^2-2abp-2bcp-2cdp+b^2+c^2+d^2\le 0\\ \Rightarrow (a^2{p}^2-2abp+b^2)+(b^2{p}^2-2bcp+c^2)+(c^2{p}^2-2cdp++d^2)\le 0\\ \Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\le 0\end{array}$$
We know that the square of a real number can never be negative.
Since, $$a,b,c,d$$ and $$p$$ are real numbers and the basic mathematical operations i.e. addition, subtraction, multiplication and division on real numbers also results in real numbers, $$(ap-b),(bp-c),(cp-d)$$ are also real numbers and their squares can not be negative.
Thus equating the squares to zero, we have
$$\begin{array}{l}(ap-b{)}^2=0\\ \Rightarrow ap-b=0\\ \Rightarrow ap=b\end{array}$$
Further solving
$$\Rightarrow p={\displaystyle \frac{a}{b}}$$ ………………………equation (1)

Similarly,
$$\begin{array}{l}(bp-c{)}^2=0\\ \Rightarrow bp-c=0\\ \Rightarrow bp=c\end{array}$$
Further solving
$$\Rightarrow p={\displaystyle \frac{c}{b}}$$ ………………………equation (2)

And also
$$\begin{array}{l}(cp-d{)}^2=0\\ \Rightarrow cp-d=0\\ \Rightarrow cp=d\end{array}$$
Further solving
$$\Rightarrow p={\displaystyle \frac{d}{c}}$$ ………………………equation (3)

From equation (1), (2) and (3) it is clear that
$${\displaystyle \frac{b}{a}}={\displaystyle \frac{c}{b}}={\displaystyle \frac{d}{c}}$$ , which implies that $$a,b,c,d$$ are in G.P. as the numbers in the sequence have a common ratio.

Option ‘B’ is correct

Note: From the given that, the relation between $$a,b,c,d$$ is established. If the four numbers have a common ratio, then they will be in G.P., but, if the four numbers have a common difference, then they will be in A.P.