JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

A_r, r=1,2,3, \ldots, n

are

n

points on the parabola

y^2=4 x

in the first quadrant. If

A_r=\left(x_r, y_r\right)

, where

x_1, x_2, x_3, \ldots, x_n

are in GP and

x_1=1

x_2=2

, then

y_n

is equal to

View answer and explanation

Correct option: A

2^{\frac{n+1}{2}}

Option A (

2^{\frac{n+1}{2}}

) is correct because the y-coordinates form a GP with a common ratio of

\sqrt{2}

, and the nth term can be calculated.

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

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

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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}