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

## Comments

Sort by:

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

Log in to reply

got the same answer yesterday in RMO

Log in to reply

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

Log in to reply

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:

\[y-x=znd\]

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

Log in to reply

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

Log in to reply

its same as in RMO Karnataka Region!

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

Log in to reply

Did u try using quadratic formula after that?

Log in to reply

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

Log in to reply

Log in to reply

I think i got it correct.

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

(Y- x)/z

Log in to reply

Why?

Log in to reply