If a1- a2- - is a sequence for which a1 - 2- a2 - 3 and an-an-1an-2 for every natural number n - 3- Then the value of a2011 is

Given a_1=3,a_2=2,a_n=a_n 1a_n 2 ∀n∈N n≥3 a_3=32, a_4=12,a_5=13,a_6=23,a_7=2,a_8=3 We have blocks of 6 numbers which are repeating. As 2011=335×6+1, we hav ...

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Byju's Answer

Standard XII

Mathematics

One - One function

If a1, a2, .....

Question

If $a_{1}, a_{2}, . . . . .$ is a sequence for which $a_{1} = 2, a_{2} = 3$ and $a_{n} = \frac{a_{n - 1}}{a_{n - 2}}$ for every natural number $n \geq 3$ . Then the value of $a_{2011}$ is

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3/2

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2/3

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Solution

The correct option is D
2

Given

$a_{1} = 3, a_{2} = 2, a_{n} = \frac{a_{n - 1}}{a_{n - 2}} \forall n \in N n \geq 3$

$a_{3} = \frac{3}{2}, a_{4} = \frac{1}{2}, a_{5} = \frac{1}{3}, a_{6} = \frac{2}{3}, a_{7} = 2, a_{8} = 3$

We have blocks of $6$ numbers which are repeating. As $2011 = 335 \times 6 + 1,$ we have $335$ blocks and one more term.

$∴ a_{2011} = 2.$

Thus sequence is $2, 3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, 2, 3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, 2, 3, \dots \dots$

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If a1,a2,..... is a sequence for which a1=2,a2=3 and an=an−1an−2 for every natural number n≥3. Then the value of a2011 is

The correct option is D 2 Given a1=3,a2=2,an=an−1an−2 ∀n∈N n≥3 a3=32,a4=12,a5=13,a6=23,a7=2,a8=3 We have blocks of 6 numbers which are repeating. As 2011=335×6+1, we have 335 blocks and one more term. ∴a2011=2. Thus sequence is 2,3,32,12,13,23,2,3,32,12,13,23,2,3,⋯⋯

If $a_{1}, a_{2}, . . . . .$ is a sequence for which $a_{1} = 2, a_{2} = 3$ and $a_{n} = \frac{a_{n - 1}}{a_{n - 2}}$ for every natural number $n \geq 3$ . Then the value of $a_{2011}$ is