Arithmetic sequence

first 3 terms of arithmetic sequence where the 21st term is -38 and the 50th term is -125

a+20d=-38

a+49d=-125

now solve simultaneously.

first 3 terms of arithmetic sequence

where the 21st term is -38 and the 50th term is -125

Let
$\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=1 & t_z=t_1= ? \\ \end{array}$

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{1} &=& (-38)\cdot (\frac{1-50}{21-50}) + (-125)\cdot (\frac{21-1}{21-50}) \\ &t_{1} &=& (-38)\cdot (\frac{-49}{-29}) + (-125)\cdot (\frac{20}{-29}) \\ &t_{1} &=& -38\cdot (\frac{49}{29}) + 125\cdot (\frac{20}{29}) \\ &t_{1} &=& \frac{-38\cdot 49+ 125\cdot 20}{29} \\ &t_{1} &=& \frac{638}{29} \\ & \mathbf{ t_{1} } & \mathbf{=} & \mathbf{22} \\ \hline \end{array}$

Let
$\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=2 & t_z=t_2= ? \\ \end{array}$

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{2} &=& (-38)\cdot (\frac{2-50}{21-50}) + (-125)\cdot (\frac{21-2}{21-50}) \\ &t_{2} &=& (-38)\cdot (\frac{-48}{-29}) + (-125)\cdot (\frac{19}{-29}) \\ &t_{2} &=& -38\cdot (\frac{48}{29}) + 125\cdot (\frac{19}{29}) \\ &t_{2} &=& \frac{-38\cdot 48+ 125\cdot 19}{29} \\ &t_{2} &=& \frac{551}{29} \\ & \mathbf{ t_{2} } & \mathbf{=} & \mathbf{19} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline d&=&t_2-t_1 \\ d&=&19-22 \\ d&=&-3 \\\\ t_3 &=& t_2 +d \\ t_3 &=& 19-3 \\ \mathbf{ t_{3} } & \mathbf{=} & \mathbf{16} \\ \hline \end{array}$

The first 3 terms of arithmetic sequence: 22, 19, 16, ...