Consider the sequence of numbers a

(1)-100.33 (2)-30.00 (3)62.33 (4)119.33

Solution follows here:

_{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}. What is the sum of first 6002 terms of this sequence?(1)-100.33 (2)-30.00 (3)62.33 (4)119.33

Solution follows here:

__Solution:__

a

_{j}= a_{j-1}– a_{j-2}_{ }=> a_{3}= a_{2}– a_{1};
a

_{4}= a_{3}– a_{2 }= a_{2}– a_{1 }– a_{2 }= -a_{1};_{}
a

_{5}= a_{4}– a_{3 }= -a_{1}– (a_{2 }– a_{1})_{ }= -a_{2};
a

_{6}= a_{5}– a_{4 }= -a_{2}– (_{ }– a_{1})_{ }= a_{1}-a_{2};
a

_{7}= a_{6}– a_{5 }= a_{1}-a_{2}– (_{ }–a_{2})_{ }= a_{1};
As a

_{7}is being equal to a_{1}, from a_{7}onwards, the series is starting to repeat.
=> First six terms are getting repeated

a

_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}= a_{1}+a_{2}+( a_{2}– a_{1})+( -a_{1})+( -a_{2})+( a_{1}-a_{2}) = 0
Sum of first six terms is zero => sum of next
six terms is also zero…

So sum of first 6000 terms is also zero, as 6000
is a multiple of six.

Sum
of first 6002 terms = (sum of first 6000 terms)+( sum of first 2 terms)

= 0+( a

Answer (3)_{1}+a_{2}) = 81.33-19 = 62.33
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