RMO 2014

Algebra Level 5

Let { a 1 , a 2 , a 3 , . . . . . . . . . , a 2016 , . . . . } \left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },.........,{ a }_{ 2016 },.... \right\} be an arithmetic progression such that it has a common difference d d and

  1. ∑ i = 1 1008 a 2 i − 1 2 = 0 \large{\sum _{ i=1 }^{ 1008 }{ { a }_{ 2i-1}^{2} } =0}
  2. ∑ i = 1 1008 a 2 i 2 = 2016 \large{\sum _{ i=1 }^{ 1008 }{ { a }_{ 2i }^{ 2 } }=2016 }

and for all k k , k ∈ Z + k\in { Z }^{ + }

a k + a k + 1 = 1 \large{{ a }_{ k }+{ a }_{ k+1 }=1}

Find the common difference d d .

This question is a numerical version of a generalized problem that appeared in RMO 2014


The answer is 2.

Department 8 by Department 8 , 27, Israel

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1 solution

Samarth Agarwal
Oct 24, 2015

a 1 2 + a 3 2 + a 5 2 + . . . . . . . . . . . + a 2015 2 = 0 . . . . . . . . . i a 2 2 + a 4 2 + a 6 2 + . . . . . . . . . . . + a 2016 2 = 2016 . . . . . . . i i i i − i : ( a 2 + a 1 ) ( a 2 − a 1 ) + ( a 4 + a 3 ) ( a 4 − a 3 ) + . . . . . . . . . . . + ( a 2016 + a 2015 ) ( a 2016 − a 2015 ) = 2016 − 0 . . . i i i a k + a k + 1 = 1 ( g i v e n ) a k + 1 − a k = d ∴ i i i c a n b e w r i t t e n a s : d + d + . . . . . . . . . + d + d ⏟ 1008 t e r m s = 2016 d = 2016 1008 d = 2 . { a }_{ 1 }^{ 2 }+a_{ 3 }^{ 2 }{ +a }_{ 5 }^{ 2 }+...........+{ a }_{ 2015 }^{ 2 }=0\quad \quad \quad .........i\\ { a }_{ 2 }^{ 2 }{ +a }_{ 4 }^{ 2 }{ +a }_{ 6 }^{ 2 }{ +...........+a }_{ 2016 }^{ 2 }=2016\quad .......ii\\ ii-i:\\ { (a }_{ 2 }{ +a }_{ 1 }){ (a }_{ 2 }-{ a }_{ 1 })+({ a }_{ 4 }+{ a }_{ 3 })({ a }_{ 4 }-{ a }_{ 3 })+...........+({ a }_{ 2016 }+{ a }_{ 2015 })({ a }_{ 2016 }-{ a }_{ 2015 })=2016-0\quad \quad ...iii\\ { a }_{ k }+{ a }_{ k+1 }=1\quad \quad (given)\\ { a }_{ k+1 }-{ a }_{ k }=d\\ \therefore \quad iii\quad can\quad be\quad written\quad as:\\ \underbrace { d+d+.........+d+d }_{ 1008\quad terms } =2016\\ d=\frac { 2016 }{ 1008 } \\ d=\boxed { 2 } .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

That's the way to solve this! Nicely presented by Samarth.

Some points to think: 1. Can there be any sequence that can have the stated conditions? To me, it looks like no sequence can satisfy all the conditions in the question.

  1. As per the first condition, Sum of squares of Odd terms is zero means the terms are complex.

  2. The condition a k + a k + 1 = 1 a_k + a_{k+1} = 1 (data) and a k + 1 − a k = 2 a_{k+1} - a_k = 2 (from solution) contradict each other

Can anyone help in synthesizing this sequence ?

Thanks!

Raghunandana Ravindra - 5 years, 7 months ago

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I don't think that any real solution to this series exist. If terms are real then odd terms should be 0 and even terms should be 1 but it is not an A.P. neither does it satisfy ii. So, may be its a series in complex numbers. It may be that the question should say a 2 n − 1 + a 2 n = 1 a n d a 2 n − a 2 n − 1 = d { a }_{ 2n-1 }+{ a }_{ 2n }=1\quad and\quad { a }_{ 2n }-{ a }_{ 2n-1 }=d and that a1,a2,a3...... are not in any A.P. as a3-a2 or a5-a4 etc. are of no use and then we just need to find d. Then only the solution is valid in reals. Since the question is a numerical version of RMO problem there may be a little bit stating problem in this question, but I can't say anything about stating as I have not seen the original question.

Samarth Agarwal - 5 years, 7 months ago

The same I did. A fine solution

Shreyash Rai - 5 years, 6 months ago

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