Latest TCS Aptitude Question SOLUTION: IN a sequence of integers,A(n)=A(n-1)+A(n-2),where A(n) is the nth 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 all 100

Given A(1)=1;A(2)=1
So, A(3)=0
A(4)=A(3)-A(2)= -1
A(5)=A(4)-A(3)= -1
A(6)=A(5)-A(4)= 0
A(7)=A(6)-A(5)= 1 = A(1)
A(8)=A(7)-A(6)= 1 = A(2)
So,next terms can easily be predicted from the previous steps.
A(9)=A(3)=0
A(10)=A(4)= -1
A(11)=A(5)= -1
A(12)=A(6)= 0
A(13)=A(7)=A(1)=1
A(14)=A(8)=A(2)=1
So, Observing the previous steps little bit carefully , we infer that the 1st 6 terms are repeated continuously.
Now, the sum of 1st 6 terms = A(1)+A(2)+A(3)+A(4)+A(5)+A(6)=1+1+0-1-1+0=0
Now, 1000 = 6*166 + 4.
so, in the 1st 1000 terms of the given sequence ,the 1st 6 terms will be repeated 166 times giving a total sum =0, and then the result will be the sum of remaining 4 terms which are effectively the 1st 4 terms of the sequence, A(1), A(2),A(3) & A(4).
So, S(1000) = 166*0 +A(1)+A(2)+A(3)+A(4) = 0 + 1 + 1 + 0 - 1 = 1.

Hence S(1000) = 1.

It ,therefore, seems that no provided option is correct , of course if my solution is not wrong.I've checked it thoroughly as far as possible, but havn't found any mistake.I of course may overlook a serious loop-whole in my solution which any1 of u can find out.
So, an earnest request to all that please check the provided solution and post ur feedback and of course with required correction(if there is any mistake) to let us know the flawless answer and the procedure for the given problem..

Thanks to all:-) :-)

@ REEMA SHARMA :

Actually , in my opinion the problem statement should be modified little bit as mentioned below:
if A(1)=-1 {but in actual posted problem A(1)=1}
and A(2)=1
with A(n)=A(n-1)-A(n-2), for n>=3 ;
Then S(1000)= ??

If the problem statement is as above ,the solution of the given problem can be furnished as follow:

A(1)= -1
A(2)= 1
A(3)= 2
A(4)=A(3)-A(2)= 1
A(5)=A(4)-A(3)= -1
A(6)=A(5)-A(4)= -2
A(7)=A(6)-A(5)= -1 =A(1)
A(8)=A(7)-A(6)= 1 =A(2)
So,next terms can easily be predicted from the previous steps.
A(9)=A(3)=2
A(10)=A(4)= 1
A(11)=A(5)= -1
A(12)=A(6)= -2
A(13)=A(7)=A(1)=-1
A(14)=A(8)=A(2)=1
So, Observing the previous steps little bit carefully , we infer that the 1st 6 terms are repeated continuously.
Now, the sum of 1st 6 terms = A(1)+A(2)+A(3)+A(4)+A(5)+A(6)=-1+1+2+1-1-2=0
Now, 1000 = 6*166 + 4
so, in the 1st 1000 terms of the given sequence ,the 1st 6 terms will be repeated 166 times giving a total sum =0, and then the result will be the sum of remaining 4 terms which are effectively the 1st 4 terms of the sequence, A(1), A(2),A(3) & A(4).
So, S(1000) = 166*0 +{A(1)+A(2)+A(3)+A(4)} = 0+ {-1+1+2+1} = 3;

Hence S(1000) = 3.

So,again a request @REEMA SHARMA to let us know whether my little correction is actually consistent with the problem-statement as it is published in the actual source of the problem..

Kindly check the problem statement and the options provided.

There is a mistake either in the Problem Statement or in the Options provided.

In my opinion, the general formula for A(n) should be [A(n-1)- A(n-2)] instead of the given expression in the problem.But, I'm not sure. The intervention of the person posting the problem is needed in this context.

all option are incorrect !
answer is -2

becouse n>=3 means first term of series will be A(3)
now A(3)=0 first term
A(4)=A(3)-A(2)= -1 second term
A(5)=A(4)-A(3)= -1 third term
A(6)=A(5)-A(4)= 0 fourth term
A(7)=A(6)-A(5)= 1 fifth term
A(8)=A(7)-A(6)= 1 sixth term
we can see from here that 1000th term will be = A(1002)
( S(1000) is sum of 1000th terms of series A(3),A(4).....A(1002) }
we can se the sum is 0 after evry 6 terms starting from A(3)...and A(998)will be the 996th term of series ....
so sum of A(3),A(4).....+A(998) = 0

now A(999) = 0 997th term
A(1000) = -1 998th term
A(1001) = -1 999th term
A( 1002 ) = 0 1000th term

Sum of 1000 terms ={ 997th +999th +999th+1000th } terms
= 0 -1-1+0
= -2
so -2 is the right answer ......according to TCS I am wrong !!!
but TCS is wrong according to evryone ..becouse 3 can not be answer for anycost

@ REEMA SHARMA :

Refer to my two solutions which I've provided.Check them meatly and thoroughly as hard as possible.Then u can easily conclude that both the solutions are correct,i.e, when A(1)=1 and A(1)= -1.

Since there is no mistake in the two solutins posted by me then it is quite obvious that there is a printing mistake in the published source of the given problem.
Lastly ,I'll again advise u to refer to my solutions bcoz these typical problems are easily solvable if u can understand the method I adopted to arrive at the solution..Dont bother about the result just try to understand how the solution is reached because 1ce u understand u can easily solve any problem of the same category no matter what data they possess..

Thanks to all..:-) :-)

Actually i have given the same question with diiferent options ny tcs only
it was
1)0
2)-1
3)-2
4)1



ans is -2
series os 0 -1 -1 0 1 1 -1 -1 .........repeats itself after every 6th term
therefore, addition o first six terms is 0
s(1000)=6*166+4
=0*166+(0+-1+-1+0)
=-2
:)

yeah right..expression is A(n-1)-A(n-2)..plz give the solution now..sorry for the mistake..

if we take A(n)=A(n-1)-A(n-2) then S(1000)=
a(n) for 3,4,5,...1000=a(1)+a(2)+ 2-1,3-2,4-3,,,,,,998-997,999-998(a terms)
=a(2)+a(999)=1+a(999)
a1=1,a2=1,a3=0,a4=-1,a5=-1,a6=0,a7=1,a8=1,a9=0,,,,,,,so on
(all terms divisible by 3 are having value as 0)
so a999=0
**** so ans will be 1+0=1

@SDEEP BHATTACHARYA
correct ansr is 3..bt dnt knw the logic behind it.can u help me?

@SDEEP BHATTACHATYA:
in the given statement A(1)=1 is given..not -1.wt to say now.plz help me..