JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

a_1, a_2, a_3, \ldots

are in A.P. If

\frac{a_1+a_2+a_3+\ldots+a_p}{a_1+a_2+a_3+\ldots+a_q}=\frac{p^2}{q^2}

, where

p \neq q

, then

\frac{a_6}{a_{21}}=

View answer and explanation

Correct option: B

\frac{11}{41}

Option B (

\frac{{11}}{{41}}

) is correct because by using the formula for the sum of an arithmetic progression and comparing ratios, we can find the ratio of the 6th term to the 21st term.

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