Latest TCS Aptitude Question SOLUTION: In a sequence of integer,A(n)=A(n-1)-A(n-2),where n is the nth term of 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 first 1000 t

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In a sequence of integer,A(n)=A(n-1)-A(n-2),where n is the nth term of 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 first 1000 terms

Asked In TCS prashant (11 years ago)

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GEETHA  (11 years ago)

given A(n)=A(n-1)-A(n-2) & A(1)=1 A(2)=1
now A(3)=A(3-1)+A(3-2)=1-1=0
similarly, A(4)=0-1=-1, A(5)=-1-0=-1, A(6)=-1+1=0..
from the 7th term the same sequence repeats.
so, first we cacl sum of first 6 tems 1+1+0-1-1+0=0
now we have to find sum of 1000 terms, for this, we divide 1000/6, there will be 4 terms left, the remaining 996 terms sum is 0. those 4 terms are 1,1,0,-1.
now sum of these 4 terms is 1. so, the ans is 1.

Is this solution Helpfull? Yes (51) | No (3)  

ayush gupta  (11 years ago)

answer is 3

Is this solution Helpfull? Yes (3) | No (12)  

Honey  (11 years ago)

Ans=1
sum of first 10 terms = 1
sum of second 10 terms = 1
sum of third 10 terms = -2
1000/30=33 + 10 as remainder
sum=0+1=1

Is this solution Helpfull? Yes (3) | No   

sameera saxena  (11 years ago)

ans is 1,series will b 1 1 0 -1 -1 0 1 1 0 -1 -1 0 0 1 1....so sum of first six term is 0 till 996 terms sum will be zero,remaining will be 1 1 0 -1 sum is 1

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rjv  (11 years ago)

A1=1
A2=1
A3=0
A4=-1
A5=-1
A6=0
A7=1
.
.
.
REM(1000/6)=4
i.e. SUM(1000)=SUM(4terms)
sum(1000)=1+1+0-1
=1(Ans)

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

geetha  (11 years ago)

a(n)=a(n-1)+a(n-2)
so, a(3)=a(2)-a(1)
a(4)=a(3)-a(2)
a(1000)=a(999)-a(1)
also s(4)=a(3)-a(1)
s(5)=a(4)-a(1)
so
s(1000)=a(999)-a(1) as remaining terms gets cancelled
now a(999)=0 as 999/6 will give reminder as 3. then s(1000)=0-1=-1.

Is this solution Helpfull? Yes | No (4)  

Arvind Kumar Yadav  (11 years ago)

s(1000)=s(999)-s(998)
=s(998)-s(997)-s(998)
=-s(997)=-(s(996)-s(995))
finaly we got s(1) which will become to 1

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