Latest TCS Aptitude Question SOLUTION: In a Sequence of A(n)=A(n-1)-A(n-2) where A(n) is the nTH term in the sequence n is the integer and n>=3,A(1)=1,A(2)=1 Calculate S(1000),where S(1000) is the sum of first 1000 term

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In a Sequence of A(n)=A(n-1)-A(n-2) where A(n) is the nTH term in the sequence n is the integer and n>=3,A(1)=1,A(2)=1 Calculate S(1000),where S(1000) is the sum of first 1000 terms?

Asked In TCS (10 years ago)

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RAKESH  (10 years ago)

A(n)=A(n-1)-A(n-2)
A(1)=1
A(2)=1
A(3)=A(3-1)-A(3-2)=A2-A1=1-1=0
A(4)=A3-A2=0-1=-1
A(5)=A4-A3=-1-0=-1
A(6)=A5-A4=-1-(-1)=0
A(7)=A6-A5=0-(-1)=1
A(8)=A7-A6=1-0=1
A(9)=A8-A7=1-1=0
A(10)=A9-A8=0-1=-1 .... so on

(1,1,0,-1,-1,0),(1,1,0,-1,...)
we see that series is repeating after 1st 6 terms

sum of 6 terms=1+1+0-1-1+0=0

1000=6*166+4=996+4


S(1000)=A(1)+A(2)+A(3)+....+A(1000)
=sum of ist 996 terms + last 4 terms
=(1+1+0-1-1+0)+(1+1+0-1-1+0)+...166 times +(1+1+0-1)
= 0+0+0+...166 times + (1+1+0-1)
=1

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sri meher  (10 years ago)

answer should be 1..pateern repeats for evry six termsi.e1,1,0,-1,-1,0..sum of first 6 terms is 0..up to 996 sum becomes zero..hence frm 996 to 1000..sum equal to 1+1+0+-1=1..therefore ans is 1..thank u..

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Deepali Khera  (10 years ago)

if you solve the expression A(n)=A(n-1)-A(n-2)
put n= 3 you get, A(3)=A(2)-A(1)=1-1=0
thus now put n= 4... you will get series as 1,1,0,-1,-1,0,1,1,0,-1...
here S(10)i.e. sum of 10 terms = 1
thus sum of 1000 terms will be 1000

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prem kumar  (9 years ago)

f(3)=1 and f(4)=-1
so there cancel each other
so on........f(999)=1 and f(1000)=-1 cancel each other
ans is 0

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