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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

Let

{a_1},\,{a_2},\,....\,,\,{a_{10}}

be a GP. If

\frac{{{a_3}}}{{{a_1}}} = 25,

then

\frac{{{a_9}}}{{{a_5}}}

equals

View answer and explanation

Correct option: D

{5^4}

Option D (

{5^4}

) is correct because the ratio

\frac{{{a_9}}}{{{a_5}}}

simplifies to

{r^4}

, and given

\frac{{{a_3}}}{{{a_1}}} = 25

, we find

{r^4} = {5^4}

.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

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

, then find the following
(i)

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

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

are in G.P. and

b,y,c

are in G.P., then

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

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

is expressed as a polynomial in

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

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