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 \[\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}\].
If \[a,b,c,d\] are in H.P., then \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}\] 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
\[ \Rightarrow p = \dfrac{a}{b}\] ………………………equation (1)

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

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

From equation (1), (2) and (3) it is clear that
\[\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}\] , 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.