IIT JEE , AIEEE Forum - Mathematics , Algebra

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OK , HERE IS THE THIRD ONE......................

In a finite sequence of real numbers the sum of any seven successive terms is

negative, and the sum of any eleven successive terms is positive. Determine the

maximum number of terms in the sequence.

AWARD : 20 RATING POINTS

REQUEST - THIS POST IS NOT FOR PEOPLE WHO HAVE SOLVED THIS Q. BEFORE OR HAVE AN IDEA FROM WHERE TO GET THE SOLUTION ( BY SOME UNKNOWN REFERENCES .............. )

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I PREDICT THIS SHOULD B SOLVED BY MANY.............................

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C'MON EVERYBODY........

I'M EXPECTING SOME REPLIES YAAR.............

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I AM WAITING........................

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IS EVERYBODY'S BULB FUSED AT THE MOMENT...............

THIS ISNT THAT DIFFICULT AS THE PREVIOUS ONE.......

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OK - A HINT ( NOT A VER USEFUL ONE )

THE NO OF TERMS IN THIS SEQUECE IS A SUM OF 2 PERFECT CUBES............

nano0101

OK

_{^{Let the series be a,a+d,a+2d .................}}

when n=7

S_{7 ^{< 0}}

_{^{7/2 ( 2a+6d) < 0}}

_^a+3d<0

_{^{means the 4th term < 0}}

_{^{when n= 11}}

_^S11>0

_{11/2(2a+10d)>0}

_a+5d>0

_{the 6th term >0}

_{It means that till 4th term evry term must be -ve n after 6th term +ve . 5th cud be anything.}

_{thus a must be some -ve number.}

_{if 5th term is = 0}

_{then max no of terms = 10}

_{if 5th term>0}

_{max no of terms =9}

_{if 5th <0}

_{max no of term= 11}

_{since those 11 cud be anywhere in the series}

_{max no of terms = 12}

_{if it is more than that the fourth term will become more than 0}

_{thus answer =12}

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I DIDNT SAY IT WAS AN AP

IT CAN BE ANY RANDOM SERIES OF REAL NOS.

SORRY , INCORRECT ANSER .

ALSO , 11 IS NOT A SUM OF 2 PERFECT CUBES AS GIVEN IN MY HINT..............

nano0101

wow atlast the answer came!

good work amit!

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any experts who wud like to solve this.................

anit_sahu

is the answer 16

if correct will submit solution later in evening

fused_bulb

fantastic anit_sahu

hope u did it without any aid.........

looking forward for the proof and the 20 points will be urs.

congrats...............

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any other views..........

anit_sahu

we first show that such a sequence cant have more than 17 or more numbers

let us assume that the numbers are a1,a2,a3-----a17

a1 a2 a3 a4 a5 a6 a7

a2 a3 a4 a5 a6 a7 a8

a3 a4 a5 a6 a7 a8 a9

----------------------------

a11 a12 a13 a14 a15 a16 a17

the sum of the terms in positive in every column and negative in every row

if by adding all rows it is positive and adding all columns it is negative

which is a contradiction

so we can have at most 16

terms

which is poosible by 8,8,-21,8,8,8,-21,8,8,-21,8,8,8,-21,8,8

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