Q.Facing the Pandemic (The Virus Spread): The population of a certain city is 51,20,000. Assume 5000 persons in the city are infected by corona-virus. The virus is so contagious that the number of infected persons doubles every 12 days. If the rate of doubling remains constant, then in how many days will 50% of the population get infected?

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. …