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

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40022343) TCS question aptitude Keep an EYE

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

Asked In TCS Deborah (10 years ago)

Solved Deborah Read Solution (10)

Is this Puzzle helpful?   (12)   (7) Submit Your Solution

Read Solution (10) :

Deborah  (10 years ago)

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
for suppose if we add first ten terms we get result as 1 i.e 1+1+0-1-1+0+1+1+0-1=1
hence S(1000)=1

Is this solution Helpfull? Yes (43) | No (2)  

Uma  (10 years ago)

ans is 1.
a(1)=1
a(2)=1
a(3)=0
a(4)=-1
a(5)=-1
a(6)=0
{1,1,0,-1,-1,0}repeats for 166 times i.e upto 996th term and it's sum equal to zero.now the sum of last four terms(1,1,0,-1)=1.
sum(1000 terms)=1.

Is this solution Helpfull? Yes (18) | No (1)  

mayank kumar  (10 years ago)

A(1)=1,A(2)=1,A(3)=1-1=0,A(4)=0-1=-1,A(5)=-1-0=-1,A(6)=-1-(-1)=0,A(7)=0-(-1)=1,A(8)=1-0=1,A(9)=1-1=0
hence we can see that term multiple of 3 becomes 0....
now,
S(from n=3 to 1000)=A(3)+A(4)+A(5)+.......+A(999)+A(1000)
=A(2)-A(1)+A(3)-A(2)+A(4)-A(3)+.......+A(998)-A(997)+A(999)-A(998)
=A(999)-A(1)=-1(SINCE A(999)=0)
hence,s(from n=1 to 1000)=A(1)+A(2)+S(from n=3 TO 1000)=1+1-1=1(ans)

Is this solution Helpfull? Yes (4) | No   

Jeyapreetha  (10 years ago)

A(1)=1,A(2)=1.
A(3)=A(2)-A(1) acc to the given formula A(n)=A(n-1)-A(n-2)
A(3)=1-1=0.
similarliy,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.
after 6th term seq is repeating like...1,1,0,-1,-1,0...after every 6 terms
so the sum of these 6 terms are 1+1+0+(-1)+(-1)+0=0.
now we have to find sum of 1000 terms...
sum of first 600 terms is 0...(as consider 1,1,0,-1,-1,0 as one sequence,now consider 166 seq i.e 166*6=996 terms)
so the sum of first 996 terms is 0...after that 4 terms come i.e 1,1,0,-1
so the sum of last 4 terms = 1+1+0+(-1)=1.

Is this solution Helpfull? Yes (3) | No (1)  

SANJIB SADHUKHAN  (10 years ago)

A1=1
A2=1
A3=A1-A2=0
A4=-1
A5=-1
A6=0
A7=1
A8=1
.
.
A999=-1
A1000=-1
S1000=1+1-A1+A999=2-1-1=0

Is this solution Helpfull? Yes (2) | No   

nagashree  (10 years ago)

@debs agatle kada :P :P

Is this solution Helpfull? Yes (2) | No (1)  

Ratnesh srivastava  (10 years ago)

s(1000)=a[1]+a[2]+a[3]+a[4]....+a[1000]
=1+1+0-1-1++0+1+1+0-1 10 terms

Is this solution Helpfull? Yes (1) | No (1)  

priyam  (10 years ago)

A(3)=0
A(4)=0-1=-1
A(5)=-1-0=-1
A(6)=0
A(7)=0--1=1
A(8)=0--1=1
A(9)=1-1=0
A(10)=0-1=-1
S(1000)=0

Is this solution Helpfull? Yes (1) | No (2)  

pranay sharma  (10 years ago)

ans is 1 .
a(1000)=a(1000-1)-a(1000-2)
a(1000)=1

Is this solution Helpfull? Yes | No (2)  

Arun Menariya  (10 years ago)

ANS IS 2;
TERMS ARE 1,2,0,-1,-1,0,1,1,0,-1,-1,0,1,1,-1,-1.........
SUM OF FIRST TWO TERMS IS 3;
SUM OF NEXT 6 TERM IS 0;
AFTER ELEMINATING FIRST TWO TERMS THERE ARE 98 REMAINING TERMS OUT OF WHICH 96 TERM GIVE 0 SUM AND LAST TWO GIVE-1(0,-1 terms) so sum is 3-1=2

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.