Previous Year Paper
CAT2004Previous Years Paper
Consider the sequence of numbers a_{1}, a_{2}, a_{3}…to infinity where a_{1}= 81.33 and a_{2}= −19 and a_{j} = a_{j}_{}_{1}– a_{j}_{}_{2}for 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

Given a_{1}= 81.33; a_{2}= − 19
Also:
aj = a_{j}_{}_{1}– a_{j}_{}_{2}, for j ≥ 3
⇒ a_{3}= a_{2}− a_{1}= −100.33
a_{4}= a_{3}− a_{2}= −81.33
a_{5}= a_{4}− a_{3}= 19
a_{6}= a_{5}− a_{4}= +100.33
a_{7}= a_{6}− a_{5}= +81.33
a_{8}= a_{7}− a_{6}= −19
Clearly onwards there is a cycle of 6 and the sum of terms in every such cycle = 0. Therefore, when we add a_{1}, a_{2}, a_{3}... upto ‘2’ we will eventually be left with a_{1}+ a_{2}only i.e., 81.33 − 19 = 62. 33.
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