Latest TCS Aptitude Question SOLUTION: In a sequence of integers, A(n) = A(n-1)-A(n-2).where A(n) is nth term in sequence. n is integer and n>=3.if A(1)=1, A(2)=1.calculate S(1000), where S(1000) is sum of first 1000 te

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In a sequence of integers, A(n) = A(n-1)-A(n-2).where A(n) is nth term in sequence. n is integer and n>=3.if A(1)=1, A(2)=1.calculate S(1000), where S(1000) is sum of first 1000 terms.

a)2 b)3 c)4 d)0

Asked In TCS prerna rai (11 years ago)

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Debarati Dey  (11 years ago)

Given:
A(1)=1
A(2)=1

A(n)=A(n-1)-A(n-2) & n>=3

THUS,
A(3)= A(2) -A(1)= 1-1= 0
A(4)= A(3) -A(2)= 0-1=-1
A(5)= A(4) -A(3)= -1-0=-1
A(6)= A(5) -A(4)= -1+1= 0
A(7)= A(6) -A(5)= 0+1= 1
A(8)= A(7) -A(6)= 1-0= 1
A(9)= A(8) -A(7)= 1-1= 0
A(10)=A(9) -A(8)= 0-1=-1
A(11)=A(10)-A(9)= -1-0=-1
A(12)=A(11)-A(10)=-1+1= 0
A(13)=A(12)-A(11)= 0+1= 1
.
.
.
.
this continues till A(1000).

a pattern is formed at an interval of 6 i.e.{1,1,0,-1,-1,0}
So, 1+1+0+(-1)+(-1)+0=0

since there r 1000 terms,1000/6= quotient=166,remainder=4.

thus till the 996th(=166*6) term the sum will be 0 and for the rest of the 4 terms the sum will be {1+1+0-1}=1.

So S(1000)=1

The given options are incorrect.

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vikas kale  (11 years ago)

a1=1;
a2=1;
a3=a2-a1=0;
a4=a3-a2=-1;
a5=a4-a3=-1;
a6=a5-a4=0
group ogf five no sub is 0
so till 995=0 ans was come
and a997=-1
ans=3
a998=-1

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