JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

(a, b),(c, d),(e, f)

are the vertices of a triangle such that

a, c, e

are in GP with common ratio

r

and

b, d, f

are in GP with common ratio

s

, then the area of the triangle is

View answer and explanation

Correct option: C

\frac{a b}{2}(r-1)(s-1)(s-r)

Option C (

\frac{{a b}}{{2}}(r-1)(s-1)(s-r)

) is correct because the area of the triangle can be calculated using the determinant formula with the given vertices, and simplifying the determinant leads to the expression.

More Questions from SEQUENCE OF SERIES

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Given the sets

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

, then find the following
(i)

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

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

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

are in Suppose

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

is expressed as a polynomial in

y

, as

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

where

y=x-2

, then

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

equals If

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

and

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

are in

A P

, where

a, b

and c are in

G P

, then

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