Q.The 4th term of a G.P. is square of its second term, and the first term is $-3$. Determine its 7th term.

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$.

If the first term is $a$, then the sequence looks like:

The $n$-th Term

To find any term directly without listing all previous ones, use the formula:

where $T_n$ is the $n$-th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$, $r = 2$. The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$, which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

When $r = 1$, every term is just $a$, so the sum is simply $n \times a$.

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$. Multiply both sides by $r$: $rS = ar + ar^2 + \dots + ar^n$. Subtract the first from the second: $rS - S = ar^n - a$, so $S(r-1) = a(r^n - 1)$, giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$, $r=\frac12$, so $S_\infty = \frac{1}{1 - 1/2} = 2$. This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

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