JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

a, b, c, d

and

p

are distinct real numbers such that

\left(a^2+b^2+c^2\right) p^2-2 p(a b+b c+c d)+\left(b^2+c^2+d^2\right) \leq 0

then

a, b, c, d

are in

View answer and explanation

Correct option: B

G.P

The numbers

a, b, c, d

are in G.P. because the given inequality implies that

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

, which is the condition for a geometric progression.

More Questions from SEQUENCE OF SERIES

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

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a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

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y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}