The sum of first three terms of a Geometric Progression (G.P.) is $\frac{21}{2}$ and their product is 27. Given $r \lt 1$, find its:

Let the first three terms of the G.P. be $\frac{a}{r}, a$ and $ar$.

(a) Given,

Product of the terms = 27.

$$\begin{aligned}\Rightarrow \frac{a}{r} \times a \times ar = 27 \\[1em] \Rightarrow a^3 = 27 \\[1em] \Rightarrow a = 3.\end{aligned}$$

Given,

Sum of the terms = $\frac{21}{2}$.

$$\begin{aligned}\Rightarrow \frac{a}{r} + a + ar = \frac{21}{2} \\[1em] \Rightarrow a\Big(\frac{1}{r} + 1 + r\Big) = \frac{21}{2} \\[1em] \Rightarrow 3\Big(\frac{1}{r} + 1 + r\Big) = \frac{21}{2} \\[1em] \Rightarrow \frac{1}{r} + 1 + r = \frac{7}{2} \\[1em] \Rightarrow \frac{1}{r} + r = \frac{7}{2} - 1 \\[1em] \Rightarrow \frac{1}{r} + r = \frac{5}{2}.\end{aligned}$$

Multiplying both sides by $2r$,

$$\begin{aligned}\Rightarrow 2 + 2r^2 = 5r \\[1em] \Rightarrow 2r^2 - 5r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 4r - r + 2 = 0 \\[1em] \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\[1em] \Rightarrow (2r - 1)(r - 2) = 0 \\[1em] \Rightarrow r = \frac{1}{2} \text{ or } r = 2.\end{aligned}$$

Since $r \lt 1, r = \frac{1}{2}$.

Hence, the common ratio = $\mathbf{\frac{1}{2}}$.

(b) With $a = 3$ and $r = \frac{1}{2}$,

First term = $\frac{a}{r} = \frac{3}{\frac{1}{2}} = 6$

Second term = $a = 3$

Third term = $ar = 3 \times \frac{1}{2} = \frac{3}{2}$.

Hence, the first three terms are 6, 3 and $\mathbf{\frac{3}{2}}$.