Q.A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

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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