a) Convert the tapered bar into step bar,
Let, it be converted into two elements of equal length.

$A_{1/2}=\frac{80+20}{2}=\frac{100}{2}=50mm^2$
$A_{1}=\frac{A_0+A_{1/2}}{2}=\frac{80+50}{2}=65mm^2$
$A_{2}=\frac{A_{1/2}+A_0}{2}=\frac{50+20}{2}=35mm^2$

b) Element matrix Equation is given by,

$\begin{bmatrix} K \end{bmatrix}^e=\frac{AE}{h_e}\begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix}$

$For\,\, element \,\,1,\,\,A=65mm^2 \,\,,\,\, h_e=30mm$
$\frac{AE}{h_e}=\frac{65*2.1*10^5}{30}=455*10^3$
$\begin{bmatrix} K \end{bmatrix}^1=10^3\begin{bmatrix} 455&-455\\ -455&455 \end{bmatrix}$

$For\,\, element \,\,2,\,\,A=35mm^2 \,\,,\,\, h_e=30mm$
$\frac{AE}{h_e}=\frac{35*2.1*10^5}{30}=245*10^3 N/mm $
$\begin{bmatrix} K \end{bmatrix}^2=10^3\begin{bmatrix} 245&-245\\ -245&245 \end{bmatrix}$

c) Elemental matrix
$\begin{bmatrix} K \end{bmatrix}=10^3\begin{bmatrix} 455&-455&0\\-455&455+245&-245\\0&-245&245 \end{bmatrix}$
$ \hspace{2cm}=10^3\begin{bmatrix} 455&-455&0\\-455&700&-245\\0&-245&245 \end{bmatrix}$

d) Global Matrix Equation,
$\begin{bmatrix} K \end{bmatrix}\begin{Bmatrix} u \end{Bmatrix}=\begin{Bmatrix} P \end{Bmatrix}$
$ 10^3\begin{bmatrix} 455&-455&0\\-455&700&-245\\0&-245&245 \end{bmatrix}\begin{Bmatrix} u_1\\ u_2 \\u_3\end{Bmatrix}=\begin{Bmatrix} P_1 \\ P_2\\P_3 \end{Bmatrix}$

e) Imposing boundary condition,
$u_1=0\,\,,\,\,P_3=500N$
$P_2=0\hspace{3cm}$ -for balancing,

$ 10^3\begin{bmatrix} 455&-455&0\\-455&700&-245\\0&-245&245 \end{bmatrix}\begin{Bmatrix} 0\\ u_2 \\u_3\end{Bmatrix}=\begin{Bmatrix} P_1 \\ 0\\500 \end{Bmatrix}$

f) Frame the equation,
$10^3(-455*u_2)P_1\hspace{9cm}$ -(1)
$10^3(700u_2-245u_3)=0\hspace{8cm}$ -(2)
$10^3(-245u_2+245u_3)=500\hspace{7cm}$ -(3)

Solving equation (2) and (3) we get,
$\therefore \,\,\,u_2=0.0011mm\hspace{2cm}u_3=0.00314mm$
Substituting in equation (1), we get,
$P_1=-500N$
$\sum P_x=P_1+P_2+P_3=-500+0+500=0 \hspace{3cm} $ -verified