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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is

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Correct option: D

2 + \sqrt 3

Option D (

2 + \sqrt 3

) is correct because solving the quadratic equation derived from the AP condition yields this value for the common ratio, considering the GP is increasing.

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

, then find the following
(i)

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

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

are in G.P. and

b,y,c

are in G.P., then

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

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

is expressed as a polynomial in

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

are the lengths of sides of If the sum of three numbers in G.P. is 63 and the product of the first and the second term is

\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