If $3,3^2, 3^3$ are in $G P$, then $\log 3, \log 9$ and $\log 27$ are in

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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3,3^2, 3^3

are in

G P

, then

\log 3, \log 9

and

\log 27

are in

View answer and explanation

Correct option: A

A P

Option A (

A P

) is correct because the logarithms of terms in a geometric progression form an arithmetic progression.

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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}