JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

Let

a_{1}, a_{2}, a_{3}, \ldots

be an A.P. If

a_{7}=3

, the product

a_{1} a_{4}

is minimum and the sum of its first

n

terms is zero, then

n !-4 a_{n}(n+2)

is equal to :

View answer and explanation

Correct option: B

The correct answer is 24 because after finding the values of a and d, and subsequently n, the expression n! - 4*a_n*(n+2) evaluates to 24.

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

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

A-(B \cap C)

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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

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

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

A P

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G P

a, b, c

\frac{3}{4}