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Numerical Ability
Co-ordinate geometry
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
a)2
b)3
c)4
d)0
Read Solution (Total 17)
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- actually answer is 1...
consider the soln...
given A(1)=1,A(2)=1.
so, 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...now consider in splitted pattern,like
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.
so according to me 1 is answer....
plzzz give ur feedback - 12 years agoHelpfull: Yes(264) No(22)
- Ans. 1 (Not in Option)
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
And so on.... we can clearly see that After every n/6 sum of n no become 0
So till S(996) total will be 0
A(997)=1
A(998)=1
A(999)=0
A(1000)=-1
So S(1000) = 1 - 12 years agoHelpfull: Yes(23) No(3)
- answer is 0.
given A(1)=1,A(2)=1.
so, 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...
996th term is 6th term(due to repetetion),1000 th term is 4th one.i.e..-1.
it is in A.P.s(n)=(n/2)(a+l)where l is last term,a is 1st term.
(1000/2)(1-1)=0.
k....0 is tje answer. - 12 years agoHelpfull: Yes(13) No(26)
- actual answer should 1 according to math calculation but according to open seesame answer will be 3......
how it would be i dnt knw...... - 12 years agoHelpfull: Yes(10) No(4)
- @ divyank and aashly.....we have to find sum of first 1000 terms instead of 1000th term.....read question again carefully....
- 12 years agoHelpfull: Yes(9) No(7)
- if u calculate A(8)=0
A(16)=0
..
..
..
A(1000)=A(8*125)=0
as the series is repeating - 12 years agoHelpfull: Yes(8) No(30)
- trial and error
now 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
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.
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A(1000)=0
so sum is 0 - 12 years agoHelpfull: Yes(7) No(37)
- but 1 is not even in one of the options of openseesame test of tcs.
- 12 years agoHelpfull: Yes(7) No(3)
- finally tell any one answers yar... so many confusions...
- 12 years agoHelpfull: Yes(5) No(6)
- ans is 1.
sum of first 6 terms is 0
sum of next 6 terms is 0
likewise
1000%6 = 4
last four terms are 1,1,0,-1
hence ans is=1. - 11 years agoHelpfull: Yes(3) No(0)
- -1
A(1000)=A(1999)-A(1998)
A(1999)=A(1998)-A(1997)
.
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A(4)=A(3)-A(2)
A(3)=A(2)-A(1)............sum up all terms
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A(1000)=-A(1)
=-1(ans) - 12 years agoHelpfull: Yes(2) No(7)
- here,,it is mentioned that no. of terms are 1000 whose sum have to be calculated and n>=3 so terms will start from 3 to 1003 instead of 1->1000
hence, 0,-1,-1,0,1,1 so sum of 6 terms is 0
therefore..sum of first 996 terms will be 0 as 996 is divisible by 6
now last 4 terms are 0,-1,-1,0
sum = -2 - 11 years agoHelpfull: Yes(2) No(2)
- the answer of this question should be -1
also the options in the question are wrong - 10 years agoHelpfull: Yes(0) No(0)
- actually answer is 0...
consider the soln...
given A(1)=1,A(2)=1.
so, 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.
A(7)=0-(-1)=1
A(8)=1-0=1
after 6th term seq is repeating like...0,-1,-1,0,1,1,0,-1,-1,0,1,1,0,-1...and so on....
REMEMBER, n>=3 ...so n starts from 3. NOT from 1. so the 1000th term is A(1003)
so now the sum of the first 6 terms are 0+(-1)+(-1)+0+1+1=0.
So, sum of 996 terms is also 0. so A(999)=0
now the next 4 numbers in the series are-- 0,-1.-1.0
so the sum upto A(1003) i.e. 1000 terms starting from n=3 is (-2) - 9 years agoHelpfull: Yes(0) No(0)
- We have A(n) = A(n-1)-A(n-2)
also given are A(1)=A(2)=1
so, A(3) will come as 0
A(4)=-1
A(5)=-1
and A(6)= 0
Sum of first 6terms, that is from A(1)+A(2)+...+A(6)= 1+1+0-1-1+0=0
SO, SUM OF FIRST 600 TERMS = 0
SO SUM OF FIRST 996 TERMS (as 996 is the last number before 1000 that is a multiple of 6 ) = 0
Now repeating the cycle we get ,
A(997)=1
A(998)=1
A(999)=0
A(1000)=-1
NOW...
A(1000) = A(999) - A(998)
A(999) = A(998) - A(997)
A (997) = A(996)-A(995) NOTICE THAT A(998), A(997), A(996),A(995) AND SO ON WILL GET CANCELLED
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.
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A(5) = A(4) - A(3)
A(4)= A(3)-A(2)
A(3)=A(2)-A(1)
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WE SEE AT THE END OF THE SERIES OF SUM WE HAVE ONLY A(999)-A(1) AS THE REST OF THE NUMBERS GET CANCELLED.
WE KNOW A(999)= 0 AND A(1) = 1
SO A(999)-A(1) = 0-1 = -1 ANSWER.
PLEASE GIVE FEEDBACK GUYS!! - 9 years agoHelpfull: Yes(0) No(1)
- for every completion of 6 terms the series is repeating so, the greatest 3 digit number that is divisible by 6 is 996
i.e., upto 996 the sum =0
rest of the series is 1,1,0,-1
so, sum=1+1+0-1=1
But answer is not found in the option. - 9 years agoHelpfull: Yes(0) No(0)
- ans:1
not in option - 7 years agoHelpfull: Yes(0) No(0)
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