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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

Let the first term

a

and the common ratio

r

of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033 , then the sum of these three terms is equal to

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Correct option: B

231

Option B (231) is correct because by factoring the given sum of squares and solving for the first term and common ratio, the sum of the first three terms of the GP is found to be 231.

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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