JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+

\frac{1}{(180-a)(200-a)}=\frac{1}{256}

, then the maximum value of

a

is :

View answer and explanation

Correct option: C

212

Option C (212) is correct because solving the given equation leads to a quadratic equation in 'a', and the maximum value of 'a' is 212.

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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

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y=x-2

\sum\limits_{{i=0}}^{2012} a_i

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\log \left(\frac{a}{3 b}\right)

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G P

a, b, c

\frac{3}{4}