JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Understanding +4 | -1

Assertion: Sum to

n

terms of the geometric progression

x^3, x^5, x^7, \ldots(x \neq \pm 1) \text { is } \frac{x^3\left(1-x^{2 n}\right)}{\left(1-x^2\right)} .

Reason: If '

a

' is the first term and

{r}

is common ratio of a G.P. then sum to

{n}

terms is given as

{S}_{{n}}=\frac{{a}\left({r}^{{n}}-1\right)}{{r}-1} \text { or }=\frac{{a}\left(1-{r}^{{n}}\right)}{1-{r}} \text { if } {r} \neq 1 .

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct explanation for assertion

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the sum to n terms of the given G.P. is correctly calculated using the formula provided in the reason.

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