Given that, $a, b$ and $c$ are in GP.

Then, $$\frac{b}{a}=\frac{c}{b}=r$$ [constant]

$$\begin{aligned} \Rightarrow \quad & b =a r \Rightarrow c=b r \\ \therefore \quad & \frac{a-b}{b-c} =\frac{a-a r}{a r-b r}=\frac{a(1-r)}{r(a-b)}=\frac{a(1-r)}{r(a-a r)} \\ & =\frac{a(1-r)}{a r(1-r)}=\frac{1}{r} \\ \therefore \quad & \frac{a-b}{b-c} =\frac{1}{r}=\frac{a}{b} \text { or } \frac{b}{c} \end{aligned}$$

Let AP be $a, a+d, a+2 d \cdots a+(n-1) d$

$$\begin{array}{l} \therefore \quad a_1+a_n & =a+a+(n-1) d \\ & =2 a+(n-1) d \quad \text{.... (i)}\\ \text { Now, } \quad a_2+a_{n-1} & =(a+d)+[a+(n-2) d] \\ & =2 a+(n-1) d \\ & a_2+a_{n-1} =a_1+a_n \quad \text{[using Eq. (i)}\\ & a_3+a_{n-2} =(a+2 d)+[a+(n-3) d] \\ & =2 a+(n-1) d \\ & =a_1+a_n \quad \text{[using Eq. (i)} \end{array}$$

Follow this pattern, we see that the sum of terms equidistant from the beginning and end in an AP is equal to [first term + last term].

It is given that, $T_3=4$

Let $a$ and $r$ the first term and common ration, respectively.

$$\begin{aligned} \text{Then, }\quad a r^2 & =4 \quad \text{.... (i)}\\ & \text { Product of first } 5 \text { terms }=a r \cdot a r \cdot a r^2 \cdot a r^3 \cdot a r^4\\ & =a^5 r^{10}=\left(a r^2\right)^5=(4)^5\quad \text{[using Eq. (i)]} \end{aligned}$$