Hint:
Solve the equation $$\log_2X-\log_2(X+9)=\log_2(X+9)-\log_2(X+45)$$
for X. This follows from the fact that the difference between any two consecutive terms is the same.
So you get
$$\frac{X}{X+9}=\frac{X+9}{X+45}$$ which gives $X=3$
Observe that the common difference will be $\log_2(X+9)-\log_2X$.
You can now obtain the 5th term once you know the common difference.
Let the arithmetic progression be $\langle x_1,x_2,x_3,\dots\rangle$, so that $\alpha=x_p$ and $\beta=x_q$. Let $$S=x_1+x_2+\ldots+x_{p+q}\;,$$ the sum of the first $p+q$ terms. Let $d$ be the constant difference of this progression, so that $x_{k+1}=x_k+d$ for every $k$. Then $x_{q+1}=x_q+d$, $x_{q+2}=x_q+2d$, and in general $x_{q+k}=x_q+kd$. In particular, if we set $k=p-q$, then $x_p=x_{q+k}=x_q+kd$, i.e., $\alpha=\beta+(p-q)d$, and it follows that $$d=\frac{\alpha-\beta}{p-q}\;.$$
Now write out the sum $S$ twice, as shown below, and add:
$$\begin{array}{c}
S&=&x_1&+&x_2&+&\ldots&+&x_{p+q-1}&+&x_{p+q}\\
S&=&x_{p+q}&+&x_{p+q-1}&+&\ldots&+&x_2&+&x_1\\ \hline
\end{array}$$
On the left you get $2S$. Each column on the right has the form: it contains $x_k$ and $x_{p+q+1-k}$, and its sum is $x_k+x_{p+q+1-k}$. In particular, for $k=p$ we get the sum $x_p+x_{q+1}=\alpha+\beta+d$.
Each column has the same sum (why?), and there are $p+q$ columns, so
$$2S=(p+q)(\alpha+\beta+d)\;.$$
If you now put all of the pieces together properly, you’ll get the desired formula.
Best Answer
In an arithmetic progression $a,a+d,a+2d,...$, the $n$th term is $a+(n-1)d$ and the sum to $n$ terms is $\frac{n}{2}(2a+(n-1)d)$.
If the ratio of the tenth term to the thirtieth term is $\frac{1}{3}$, then $3(a+9d)=a+29d$. If the sum of the first six terms is $42$, then $3(2a+5d)=42)$.
Solve for $a$ and $d$ and then find the first and thirteenth term.