Let $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ 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:

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Sequences and Series

Let

a, b, c, d

and p be any non zero distinct real numbers such that

(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0

. Then:

Answer: (C)

Regroup the given expression as a sum of three perfect squares:

(a^{2} p^{2} - 2 ab p + b^{2}) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) = 0

, that is

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

. A sum of squares of real numbers is zero only when each is zero, so

a p = b

b p = c

and

c p = d

. Hence

\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p

, which means

a, b, c, d

are in geometric progression.

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option