Q.If for a G.P. $t_3=\dfrac13$, $t_6=\dfrac1{81}$, find $r$.

General Term of a Geometric Progression

The Intuition: What a G.P. feels like

Imagine you start with a single grain of rice on the first square of a chessboard. On the second square, you put 2 grains. On the third, 4 grains. On the fourth, 8 grains. Each time, you multiply the previous number by 2.

That is a geometric progression. The defining rule is simple: you get the next term by multiplying the current term by a fixed number (called the common ratio, usually denoted $r$). In the rice example, $r = 2$.

So if the first term is $a$, the sequence looks like:

Notice the pattern: the exponent on $r$ is always one less than the position of the term. The 1st term has $r^0$, the 2nd has $r^1$, the 3rd has $r^2$, and so on.

The Precise Statement

For a geometric progression with first term $a$ and common ratio $r$ ($r \neq 0$), the $n$th term (also called the general term) is given by:

where $n$ is a positive integer ($n = 1, 2, 3, \dots$).

Why This Formula Works

Let's verify it for the first few terms:

• $n = 1$: $T_1 = a r^{0} = a$ (correct)
• $n = 2$: $T_2 = a r^{1} = ar$ (correct)
• $n = 3$: $T_3 = a r^{2} = ar^2$ (correct)

The formula is simply a compact way of saying: "start with $a$, then multiply by $r$ exactly $(n-1)$ times."

A Common Mistake to Avoid

What If $r$ Is Negative or a Fraction? …