Let $a, b$, c be in A.P. and $|a|<1,|b|<1,|c|<1$. If $x=1+a+a^2+\ldots$ to $\infty, y=1+b+b^2+\ldots$ to $\infty$ and, $z=1+c+c^2+\ldots$ to $\infty$, then $x, y$, $z$ are in

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

Let

a, b

, c be in A.P. and

|a|<1,|b|<1,|c|<1

. If

x=1+a+a^2+\ldots

\infty, y=1+b+b^2+\ldots

\infty

and,

z=1+c+c^2+\ldots

\infty

, then

x, y

z

are in

View answer and explanation

Correct option: C

Option C (HP) is correct because if a, b, c are in AP, then 1/(1-a), 1/(1-b), 1/(1-c) are in HP.

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}