Q.Find the 6th term of a GP whose first term is 7 and common ratio is 2.

Geometric Progression — From Intuition to Precision

Imagine you drop a ball from a height of 1 metre. It bounces back to half its previous height each time. The heights of successive bounces are:

$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$

What pattern do you see? Each term is obtained by multiplying the previous term by the same fixed number — here, $\frac{1}{2}$. That is the core idea of a Geometric Progression (GP).

The Intuition

A GP is a sequence where the ratio between any two consecutive terms is constant. This constant is called the common ratio, denoted by $r$.

If you start with a first term $a$, the sequence looks like:

$a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots$

Each step multiplies by $r$. If $r > 1$, the terms grow (like population doubling). If $0 < r < 1$, the terms shrink (like radioactive decay). If $r < 0$, the terms alternate in sign.

The Precise Statement

A sequence $\{t_1, t_2, t_3, \dots\}$ is a Geometric Progression if there exists a constant $r$ such that for every $n \geq 1$:

The first term is usually called $a$, so:

This is the $n$th term formula. It lets you jump directly to any term without writing out the whole sequence.

The Sum of a GP

Sometimes you need the sum of the first $n$ terms. Let $S_n = a + ar + ar^2 + \dots + ar^{\,n-1}$.

Multiply the whole sum by $r$:

Subtract the second equation from the first:

So:

If $r = 1$, the sum is simply $S_n = n a$.

A Quick Example

Find the 10th term and the sum of the first 10 terms of the GP: $2, 6, 18, 54, \dots$

Here $a = 2$, $r = \frac{6}{2} = 3$.

10th term: $t_{10} = 2 \times 3^{9} = 2 \times 19683 = 39366$

Sum of first 10 terms:

Why It Matters

GP appears everywhere — compound interest, population growth, depreciation, half-life in physics, and even in the geometry of fractals. Once you see the pattern of constant multiplication, you are looking at a geometric progression.