JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

\sqrt{3}, A

and

\sqrt{2}

are in

A P

, then

\frac{\sqrt{3}+\sqrt{2}}{2}

is greater than or equal to

View answer and explanation

Correct option: B

\sqrt{6}

Option B (

\sqrt{6}

) is correct because, since

\sqrt{3}, A, \sqrt{2}

are in AP, A is the AM of

\sqrt{3}

and

\sqrt{2}

, and AM >= GM.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

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}