JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

a_1, a_2, a_3, \ldots

are in GP with first term

a

and common ratio

r

, then

\frac{a_1 a_2}{a_1^2-a_2^2}+\frac{a_2 a_3}{a_2^2-a_3^2}+\frac{a_3 a_4}{a_3^2-a_4^2}+\ldots+\frac{a_{n-1} a_n}{a_{n-1}^2-a_n^2}

is equal to

View answer and explanation

Correct option: B

\frac{(n-1) r}{1-r^2}

Option C (4) is correct because the given condition

a, b, c

are in AP leads to the simplification of the expression to

4abc

, thus

\lambda = 4

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

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}