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 \(x, y, z, n\).

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## Comments

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TopNewest\( \displaystyle d = \frac{y-x}{nz} \)

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got the same answer yesterday in RMO

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

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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:

\[y-x=znd\]

\[d=\dfrac{y-x}{zn}\]

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In an AP if a1+a5+a10+a15+a25=300, find sum up to 24 terms?

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its same as in RMO Karnataka Region!

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

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Did u try using quadratic formula after that?

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No! I didn't. May be the equation was wrong, Sudeep Salgia's solution looks right!

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I think i got it correct.

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Well,what is it? How many did you solve in total?

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(Y- x)/z

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Why?

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