Q.A square is drawn by joining the mid-points of the sides of a square. A third square is drawn inside the second square by joining the mid-points of the second square, and the process is continued indefinitely. If the side of the original square is 8 cm, find the sum of the areas of all the squares thus formed. [FIGURE description: outer square $ABCD$ (side 8 cm), $A$ bottom-left, $B$ bottom-right, $C$ top-right, $D$ top-left, with $E$ the midpoint of $AB$. A second square $EFGH$ is formed by joining the midpoints of $ABCD$'s sides ($E$ mid $AB$, $F$ mid $BC$, $G$ mid $CD$, $H$ mid $DA$). A third square $IJKL$ is formed by joining the midpoints of $EFGH$'s sides, and the process continues inward indefinitely.]

Infinite Geometric Progression

The intuition: what happens when you keep halving?

Imagine you start walking toward a wall. In the first step, you cover half the remaining distance. In the next step, you cover half of what's left. Then half of that. And so on.

The distances you cover in each step are:

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

This is a geometric progression — each term is the previous term multiplied by a fixed number (here, $\frac{1}{2}$). That fixed multiplier is called the common ratio, $r$.

Now ask: if you keep doing this forever, what is the total distance you cover? Intuition says you never quite reach the wall — but you get arbitrarily close. The total distance covered after $n$ steps is:

$S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n}$

This sum gets closer and closer to $1$ as $n$ grows. It never exceeds $1$, but it approaches $1$ as closely as you like. We say the infinite sum equals $1$.

That is the core idea of an infinite geometric progression: a GP that goes on forever, and whose sum (under the right condition) settles to a finite number.

The precise statement

An infinite geometric progression is a sequence of the form:

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

where $a$ is the first term and $r$ is the common ratio. The sequence never ends.

The sum of the first $n$ terms (the finite sum) is:

$S_n = a \frac{1 - r^n}{1 - r}, \quad r \neq 1$

Now, what happens as $n \to \infty$? That depends entirely on $r$.

Why $|r| < 1$ is the key

Look at $S_n = a \frac{1 - r^n}{1 - r}$. The only part that depends on $n$ is $r^n$.

• If $|r| < 1$, then $r^n \to 0$ as $n \to \infty$. So $S_n \to \frac{a}{1 - r}$.
• If $|r| > 1$, then $r^n$ grows without bound — the sum diverges.
• If $r = 1$, the formula breaks (division by zero), and the sum is $a + a + a + \dots$, which diverges. …