$$ -25 ; p ; y $$ is an arithmetic sequence and $$ p ; y ;-45 $$ is a geometric sequence. Calculate the values of $$ p $$ and $$ y $$.

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To solve this problem, we need to use the properties of arithmetic and geometric sequences.

Given:
-25, p, y is an arithmetic sequence.
p, y, -45 is a geometric sequence.

For an arithmetic sequence, the difference between consecutive terms is constant. Let's denote this common difference as d.

For a geometric sequence, the ratio between consecutive terms is constant. Let's denote this common ratio as r.

From the arithmetic sequence:
-25 + d = p
p + d = y

From the geometric sequence:
p * r = y
y * r = -45

We can solve these equations to find the values of p and y.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}-25+d=p\p+d=y\pr=y\yr=-45\end{array}ight.$$
- step1: Rewrite the expression:
  $$\left\{ \begin{array}{l}p=-25+d\p+d=y\pr=y\yr=-45\end{array}ight.$$
- step2: Substitute the value of $$p:$$
  $$\left\{ \begin{array}{l}-25+d+d=y\\left(-25+dight)r=y\yr=-45\end{array}ight.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}-25+2d=y\r\left(-25+dight)=y\yr=-45\end{array}ight.$$
- step4: Rewrite the expression:
  $$\left\{ \begin{array}{l}y=-25+2d\r\left(-25+dight)=y\yr=-45\end{array}ight.$$
- step5: Substitute the value of $$y:$$
  $$\left\{ \begin{array}{l}r\left(-25+dight)=-25+2d\\left(-25+2dight)r=-45\end{array}ight.$$
- step6: Simplify:
  $$\left\{ \begin{array}{l}r\left(-25+dight)=-25+2d\r\left(-25+2dight)=-45\end{array}ight.$$
- step7: Solve the equation:
  $$\left\{ \begin{array}{l}d=\frac{-25+25r}{r-2}\r\left(-25+2dight)=-45\end{array}ight.$$
- step8: Substitute the value of $$d:$$
  $$r\left(-25+2	imes \frac{-25+25r}{r-2}ight)=-45$$
- step9: Simplify:
  $$\frac{25r^{2}}{r-2}=-45$$
- step10: Cross multiply:
  $$25r^{2}=\left(r-2ight)\left(-45ight)$$
- step11: Simplify the equation:
  $$25r^{2}=-45\left(r-2ight)$$
- step12: Rewrite the expression:
  $$5	imes 5r^{2}=5\left(-9\left(r-2ight)ight)$$
- step13: Evaluate:
  $$5r^{2}=-9\left(r-2ight)$$
- step14: Expand the expression:
  $$5r^{2}=-9r+18$$
- step15: Move the expression to the left side:
  $$5r^{2}-\left(-9r+18ight)=0$$
- step16: Remove the parentheses:
  $$5r^{2}+9r-18=0$$
- step17: Factor the expression:
  $$\left(r+3ight)\left(5r-6ight)=0$$
- step18: Separate into possible cases:
  $$\begin{align}&r+3=0\&5r-6=0\end{align}$$
- step19: Solve the equation:
  $$\begin{align}&r=-3\&r=\frac{6}{5}\end{align}$$
- step20: Calculate:
  $$r=-3\cup r=\frac{6}{5}$$
- step21: Rearrange the terms:
  $$\left\{ \begin{array}{l}d=\frac{-25+25r}{r-2}\p=-25+d\r=-3\y=-25+2d\end{array}ight.\cup \left\{ \begin{array}{l}d=\frac{-25+25r}{r-2}\p=-25+d\r=\frac{6}{5}\y=-25+2d\end{array}ight.$$
- step22: Calculate:
  $$\left\{ \begin{array}{l}d=20\p=-5\r=-3\y=15\end{array}ight.\cup \left\{ \begin{array}{l}d=-\frac{25}{4}\p=-\frac{125}{4}\r=\frac{6}{5}\y=-\frac{75}{2}\end{array}ight.$$
- step23: Calculate:
  $$\left\{ \begin{array}{l}d=-\frac{25}{4}\p=-\frac{125}{4}\r=\frac{6}{5}\y=-\frac{75}{2}\end{array}ight.\cup \left\{ \begin{array}{l}d=20\p=-5\r=-3\y=15\end{array}ight.$$
- step24: Check the solution:
  $$\left\{ \begin{array}{l}d=-\frac{25}{4}\p=-\frac{125}{4}\r=\frac{6}{5}\y=-\frac{75}{2}\end{array}ight.\cup \left\{ \begin{array}{l}d=20\p=-5\r=-3\y=15\end{array}ight.$$
- step25: Rewrite:
  $$\left(d,p,r,yight) = \left(-\frac{25}{4},-\frac{125}{4},\frac{6}{5},-\frac{75}{2}ight)\cup \left(d,p,r,yight) = \left(20,-5,-3,15ight)$$
The solutions to the system of equations are:
1. $$ d = -\frac{25}{4} $$, $$ p = -\frac{125}{4} $$, $$ r = \frac{6}{5} $$, $$ y = -\frac{75}{2} $$
2. $$ d = 20 $$, $$ p = -5 $$, $$ r = -3 $$, $$ y = 15 $$

Therefore, the values of p and y are:
1. $$ p = -\frac{125}{4} $$ and $$ y = -\frac{75}{2} $$
2. $$ p = -5 $$ and $$ y = 15 $$
The values of $$ p $$ and $$ y $$ are: 1. $$ p = -\frac{125}{4} $$ and $$ y = -\frac{75}{2} $$ 2. $$ p = -5 $$ and $$ y = 15 $$

is an arithmetic sequence and
is a geometric sequence.
Calculate the values of and .

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