### Theory:

In this section, let us look at the general formula to find the $$n^{th}$$ term of a geometric progression.

We know that the general form of a geometric progression is $$a$$, $$ar$$, $$ar^2$$,$$...ar^{n - 1}$$ with the common ratio $$r$$.

$$\text{Term } 1 = t_1 = a \times r^0 = a \times 1 = a$$

$$\text{Term } 2 = t_2 = a \times r^1 = ar$$

$$\text{Term } 3 = t_3 = a \times r^2 = ar^2$$
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$$\text{Term } n = t_{n} = a \times r^{n - 1} = ar^{n - 1}$$
Therefore, the general form of G.P is $$ar^{n - 1}$$.
Ratio between any two consecutive terms in a G.P:

$$\frac{t_2}{t_1} = \frac{ar}{a} = r$$

$$\frac{t_3}{t_2} = \frac{ar^2}{ar} = r$$

$$\frac{t_n}{t_{n - 1}} = \frac{ar^n}{ar^{n - 1}} = \frac{ar^n}{a \times r^n \times r^{-1}} = \frac{1}{r{-1}} = r$$

Thus, the ratio between any two consecutive terms of a G.P is $$r$$.

Condition for three numbers to be in G.P:

Let the three numbers $$a$$, $$b$$, $$c$$ in G.P.

$$\text{First term }= a$$

$$\text{Second term }= b = ar$$

$$\text{Third term }= c = ar^2$$

$$ac = a \times ar^2 = a^2r^2 = (ar)^2 = b^2$$

Thus, three non-zeroes numbers are in G.P only if $$b^2 = ac$$.

Important!
1. When the product of the three consecutive terms of a G.P is given, then we can consider the terms as $$\frac{a}{r}$$, $$a$$, $$ar$$.

2. When the product of the first four terms of a G.P are given, then we can consider the terms as $$\frac{a}{r^3}$$, $$\frac{a}{r}$$, $$ar$$, $$ar^3$$.

3. When every term of a G.P is divided or multiplied by the same non-zero constant, the newly-formed sequence is also in G.P.