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CAT-2004-Previous Years Paper

Question
22 out of 35
 

Consider the sequence of numbers a1, a2, a3…to infinity where a1= 81.33 and a2= 19 and aj = aj-1aj-2for j 3. What is the sum of the first 6002 terms of this sequence?



A 100.33
B 30.00
C 0 62.33
D 119.33

Ans. C

Given a1= 81.33; a2= 19

Also:

aj = aj-1aj-2, for j 3

a3= a2 a1= 100.33

a4= a3 a2= 81.33

a5= a4 a3= 19

a6= a5 a4= +100.33

a7= a6 a5= +81.33

a8= a7 a6= 19

Clearly onwards there is a cycle of 6 and the sum of terms in every such cycle = 0. Therefore, when we add a1, a2, a3... upto ‘2’ we will eventually be left with a1+ a2only i.e., 81.33 19 = 62. 33.

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