JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

a_1, a_2, \ldots

are in HP and

f_k=\sum\limits_{r=1}^n a_r-a_k

, then

2^{\alpha_1}, 2^{\alpha_2}, 2^{\alpha_3}, 2^{\alpha_4}, \ldots

are in

\left\{ {{\text{Where, }}\,{\text{ }}{\alpha _1} = \frac{{{a_1}}}{{{f_1}}},\,\,\,{\alpha _2} = \frac{{{a_2}}}{{{f_2}}},\,\,\,{\alpha _3} = \frac{{{a_3}}}{{{f_3}}}, \ldots } \right\}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because while the reciprocals of α_1, α_2, α_3 are in AP, it doesn't necessarily imply that 2^α_1, 2^α_2, 2^α_3 are in AP, GP, or HP.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

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