The GM of roots of the equation ${x^2} - 18x + 9 = 0$ is

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

The GM of roots of the equation

{x^2} - 18x + 9 = 0

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Correct option: A

Option A (3) is correct because the geometric mean (GM) of the roots of the quadratic equation is the square root of the product of the roots, which equals 3.

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Given the sets

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

, then find the following
(i)

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

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

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

are in Suppose

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

is expressed as a polynomial in

y

, as

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

where

y=x-2

, then

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

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

and

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

are in

A P

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a, b

and c are in

G P

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a, b, c

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\frac{3}{4}

of the third term, then the numbers are If the lengths of sides of a right angled triangle are in A.P., then their ratio is