JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

p, x_1, x_2 \ldots x_i, \ldots

and

q, y_1, y_2 \ldots, y_i, \ldots

are in A.P., with common difference

a

and

b

respectively, then the centre of mean position of the points

A_i\left(x_i, y_i\right)

i=1,2 \ldots n

lies on the line

View answer and explanation

Correct option: C

b x-a y=b p-a q

Option C (

b x-a y=b p-a q

) is correct because the center of the mean position lies on this line, derived from the properties of arithmetic progressions.

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a,x,b

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

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y

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y=x-2

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

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

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\frac{3}{4}