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CAT-2005-Previous Years Paper

Question
14 out of 30
 

If a1 = 1 and an+1 − 3an + 2 = 4n for every positive integer n, then a100 equals



A 399 − 200
B 399 + 200

C 3100 − 200
D 3100 + 200

Ans. C

a1= 1, an+1 – 3an + 2 = 4n

An+1 = 3an + 4n – 2

When n = 2, then a2 = 3 + 4 – 2 = 5

When n = 3, then a3 = 3 × 5 + 4 × 2 – 2 = 21

From the options, we get an idea that an can be expressed in a combination of some power of 3 and some multiple of 100.

(a) 399 – 200; tells us that an could be: 3n-1−2 × n; but it does not fit a1 or a2 or a3

(b) 399 + 200; tells us that an could be: 3n-1+ 2 × n; again not valid for a1, a2 etc.

(c) 3100 + 200; tells 3n + 2n: again not valid.

So, (c) is the correct answer.

CAT-2005-Previous Years Paper Flashcard List

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