What is the maximum number of terms in this series?

Question

We have a series where the sum of any 7 consecutive terms is negative and the sum of any 11 consecutive terms is positive. What is the maximum number of terms in this series?

Answer 1

They tell you it is impossible, but what about 1,1,1,1,1,1,-7,1,1,1,1,1,1 13 terms?



If this were an infinite sequence, then every 11 is positive while every 7 is negative would mean that every set of 4 is positive.

so if S(Tn,Tn+1,Tn+2,Tn+3) is positive, and S(Tn,...,Tn+6) is negative then S(Tn+4,Tn+5,Tn+6) is negative but S(Tn+4,...,Tn+7) is positive because it is four numbers.

thus Tn+7 is positive and we have a contradiction for an infinite sequence.



So to find the max value we have to see where this reasoning breaks down.



At 17 terms, we can say that all sequences of 4 are positive

S1-4 = S1-11 - S5-11,

S7-10 = S7-17-S11-17

S8-11=S1-11-S1-7

S14-17=S7-17-S7-13

thus all sequences of 3 are negative and all numbers must be positive a contradiction.

so max is at most 16 and at least 13

for 16 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5

fits the bill

Derived as follows



we can show every sequence of length 4 is positive except 7-10

because it is the difference of a positive sequence of 11 minus negative sequence of 7

sequences of length 3 known to be negative are then

123 =1-7-4567 234=2-8-5678 345=3-9-6789

567=5-11-891011 678=6-12-9101112 789=7-13-10111213

8910=8-14 - 11-14 91011=9-15 - 12-15

101112 10-16 - 13-16 121314=8-14 - 8-11 131415=9-15 - 9-12 141516=10-16 - 10-13

all but 456 and 11 12 13



then the following are known to be positive, as the difference of a shown positive sequence of 4 and a known negative sequence of 3

1,2,4,5,6,8,9,11,12,13,15,16

for example 1= (1234)-(234)= ((1-11)-(5-11))-((2-8)-(5-8))

= [(1-11)-(5-11)]-[(2-8)-[(5-15)-(9-15)]]

if a,a,-b,a,a,a,-b,a,a,-b,a,a,a,-b,a,a

2a+b<0

3a+b>0

b<-2a b>-3a

5a+2b <0

b<-5/2a

8a+3b>0

b>-8/3a

Answer 2

The sum of any consecutive 7 terms is the sum of the first n+7 terms minus the sum of the first n terms, where n is any whole number.

The sum of the first n terms is:

(n/2)(2a + (n - 1)*d)

The sum of the first n+7 term is:

((n+7)/2)(2a + (n + 7 - 1)*d)

So if the sum of any 7 terms is negative, we have:

((n+7)/2)(2a + (n + 7 - 1)*d) - (n/2)(2a + (n - 1)*d) < 0

(n+7)(2a + (n + 6)*d) - n(2a + (n - 1)*d) < 0

(n+7)(2a + nd + 6d) - n(2a + nd - d) < 0

2an + dn^2 + 6dn + 14a + 7dn + 42d - 2an - dn^2 + dn < 0

14dn + 14a + 42d < 0

a + dn + 3d < 0



The sum of any consecutive 11 terms is the sum of the first n+11 terms minus the sum of the first n terms, where n is any whole number.

The sum of the first n terms is:

(n/2)(2a + (n - 1)*d)

The sum of the first n+7 term is:

((n+11)/2)(2a + (n + 11 - 1)*d)

So if the sum of any 11 terms is positive, we have:

((n+11)/2)(2a + (n + 11 - 1)*d) - (n/2)(2a + (n - 1)*d) > 0

(n+11)(2a + (n + 10)*d) - n(2a + (n - 1)*d) > 0

(n+11)(2a + nd + 10d) - n(2a + nd - d) > 0

2an + dn^2 + 10dn + 22a + 11nd + 110d - 2an - dn^2 + dn > 0

22dn + 22a + 110d > 0

a + dn + 5d > 0



So we have two inequations avobe:

a + dn + 3d < 0

a + dn + 5d > 0

This yields n < -3, which is impossible.

There is no solution.

Answer 3

Impossible question.

Proof:

Sum(a_i,i=1,7) < 0 (1)

Sum(a_i,i=1,11) > 0 (2)

From (1) and (2)

Sum(a_i,i=1,4) > 0 (3)

From (1) and (3)

Sum(a_i,i=1,3) < 0 (4)

a_4th > 0

which basically means a_i > 0

So a proof by contradiction!



Edit: I concur the answer below me is correct. An assumption I made that a_4th > 0 => a_ith > 0 is wrong when the set is finite (doesn't wort at extreme ends).