RMO 2014 Delhi Region Q.2

Let $a_1,a_2,\ldots ,a_{2n}$ be an arithematic progression of positive real numbers with common difference $d$ . Let

$\large\left\{\begin{array}{l}\displaystyle\sum^n_{i=1} a^2_{2i-1}=x\\\ \displaystyle\sum^n_{i=1} a^2_{2i}=y\\\ a_n+a_{n+1}=z\end{array}\right.$

Express $d$ in terms of $x, y, z, n$ .

You can find rest of the problems here
You can find the solutions here

#Algebra #ArithmeticProgression(AP) #AlgebraicManipulation #RMO #Olympiad

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Comments

$\displaystyle d = \frac{y-x}{nz}$

Sudeep Salgia - 6 years, 6 months ago

got the same answer yesterday in RMO

Souryajit Roy - 6 years, 6 months ago

Hi Sudeep, Any tips for studying coordinate geometry for JEE?

A Former Brilliant Member - 6 years, 6 months ago

Got the same question in Tamil Nadu's RMO. My approach was as follows:

$y-x=(a^{ 2 }_{ 2n }-a^{ 2 }_{ 1 })+(a^{ 2 }_{ 2n-2 }-a^{ 2 }_{ 3 })+...+(a^{ 2 }_{ 2 }-a^{ 2 }_{ 2n-1 })$

It is a property of an AP that the sum of equidistant terms from the middle term(s) is a constant. Note that $a_n$ and $a_{n+1}$ are the middle terms of this AP. Hence, their sum $z$ is constant for all equidistant terms.

$y-x=(a_{ 2n }+a_{ 1 })(a_{ 2n }-a_{ 1 })+(a_{ 2n-2 }+a_{ 3 })(a_{ 2n-2 }-a_{ 3 })+...+(a_{ 2 }+a_{ 2n-1 })(a_{ 2 }-a_{ 2n-1 })$

$y-x=z(a_{ 2n }-a_{ 1 }) + z(a_{ 2n-2 }-a_{ 3 }) + ... + z(a_{ 2 }-a_{ 2n-1 })$

$y-x=z(a_{ 2n }-a_{ 1 } + a_{ 2n-2 }-a_{ 3 } +a_{ 2 }-a_{ 2n-1 })$

$y-x=z ((a_{2n}-a_{2n-1}) + (a_{2n-2} - a_{2n-3}) + ... + (a_2-a_1))$

The above terms are the difference of two consecutive terms, $d$ , and there are $n$ such terms:

$y-x=znd$

$d=\dfrac{y-x}{zn}$

Raj Magesh - 6 years, 6 months ago

its same as in RMO Karnataka Region!

I got a equation with x,y,z,n,d and d^2.. Couldn't go further!

Ranjana Kasangeri - 6 years, 6 months ago

I think i got it correct.