# arithematic progressions.

• November 24th 2006, 03:09 AM
anjana
arithematic progressions.
:confused: find 'd' if sum of first 'm' term is 'xm^2+ym'?
• November 24th 2006, 05:03 AM
Soroban
Hello, anjana!

What a strange problem . . .

Quote:

Find $d$ if sum of first $m$ terms is: . $xm^2 + ym$

The sum of the first $m$ terms is: . $S_m\;=\;\frac{m}{2}\left[2a + d(m-1)\right]$
. . where $a$ is the first term and $d$ is the common difference.

So we have: . $\frac{m}{2}\left[2a + d(m-1)\right] \:=\:xm^2 + ym$
. . which equals: . $\left(\frac{d^2}{2}\right)m^2 + \left(a - \frac{d}{2}\right)m \;= \;xm^2 + ym$

Two polynomials are equal if their corresponding coefficients are equal.

. . Hence, we have: . $\frac{d^2}{2} \,=\,x\quad\Rightarrow\quad\boxed{ d \,=\,\sqrt{2x}}$

Without additional information, that's the best we can do.