Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n 1. The smallest k for which ak = 0 (a) Does not exist. (b) Is 200 (c) Is 50 (d) Is 52.

Question

Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0 (a) Does not exist. (b) Is 200 (c) Is 50 (d) Is 52.

Arun · Answer

Dear Chitra Question has some missing terms as an+! is written which is incorrect. Please check and repost this. You can also attach an image. I will be happy to help you. Regards Arun (askIITians forum expert)

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Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0 (a) Does not exist. (b) Is 200 (c) Is 50 (d) Is 52.

Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0
(a) Does not exist.
(b) Is 200
(c) Is 50
(d) Is 52.

1 Answers

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Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0 (a) Does not exist. (b) Is 200 (c) Is 50 (d) Is 52.

Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0(a) Does not exist.(b) Is 200(c) Is 50(d) Is 52.

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Let a1 = 10, a2 = 20 and define an+! = an-1 – 4/an for n > 1. The smallest k for which ak = 0
(a) Does not exist.
(b) Is 200
(c) Is 50
(d) Is 52.