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If A, B, C, D  are consecutive terms in an arithmetic progression, what is the value of $$\frac{ D^2 - A^2 } { C^2 - B^2}$$

Assume B^2 is not equal to C^2.

Jun 28, 2020

#1
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If A, B, C, D  are consecutive terms in an arithmetic progression,
what is the value of $$\dfrac{ D^2 - A^2 } { C^2 - B^2}$$
Assume $$B^2$$ is not equal to $$C^2$$.

AP:

$$\begin{array}{|rcll|} \hline A &=& a \\ B &=& a+d \\ C &=& a+2d \\ D &=& a+3d \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{ D^2 - A^2 } { C^2 - B^2}} &=& \dfrac{(D-A)(D+A)} { (C-B)(C+B)} \\\\ &=& \dfrac{(a+3d-a)(a+3d+a)} { \Big(a+2d-(a+d)\Big)(a+2d+a+d)} \\\\ &=& \dfrac{3d(2a+3d)} { (a+2d-a-d)(2a+3d)} \\\\ &=& \dfrac{3d(2a+3d)} { d(2a+3d)} \\\\ &=& \mathbf{3} \\ \hline \end{array}$$ Jun 29, 2020
#2
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Let d be the common difference.

$$\quad\dfrac{D^2 - A^2}{C^2 - B^2}\\ = \dfrac{(A + 3d)^2 - A^2}{(A + 2d)^2 - (A + d)^2}\\ = \dfrac{((A + 3d) - A)((A + 3d) + A)}{((A + 2d) - (A + d))((A + 2d) + (A + d))}\\ = \dfrac{3\color{red}d (2A + 3d)}{\color{red}d(2A+ 3d)}\\ = 3$$

Jun 29, 2020