If ${a_1} = {a^2} = 2,\;{a_n} = {a_{n - 1}} - 1\left( {n > 2} \right),\;{\rm{then}}\;{a_5}$ is

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

{a_1} = {a^2} = 2,\;{a_n} = {a_{n - 1}} - 1\left( {n > 2} \right),\;\mathrm{then}\;{a_5}

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because, given the recursive formula,

a_5

is calculated to be -1.

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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

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}