JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

Let

a_1, a_2, \ldots, a_9

be in G.P. with

a_1 < 0

such that

a_1+a_2=4, a_3+a_4=16

. If

\sum\limits_{i=1}^9 a_i=4 \lambda

, then

\lambda

is equal to

View answer and explanation

Correct option: C

-171

Option C is correct because by finding the common ratio and the first term of the G.P., the sum of the first 9 terms can be calculated, leading to

\lambda = -171

More Questions from SEQUENCE OF SERIES

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}