JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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x^2+9 y^2+25 z^2=x y z\left(\frac{15}{x}+\frac{5}{y}+\frac{3}{z}\right)

, then

x, y

and

z

are in

View answer and explanation

Correct option: D

H.P.

Option D (H.P.) is correct because the given equation implies x = 3y = 5z, which means x, y, and z are in H.P.

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)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

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

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

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

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}