Q.The 5th, 8th and 11th terms of a G.P. are $p$, $q$ and $s$, respectively. Show that $q^2 = ps$.

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.

Quick Reference Table

Common Mistakes to Avoid

• Confusing $n$ and $n-1$: The first term corresponds to $n=1$, so the exponent is $n-1$, not $n$.
• Using infinite sum when $|r| \geq 1$: The formula gives a finite number, but the actual sum is infinite — it's a trap.
• Forgetting the sign when $r$ is negative: Terms alternate, and the sum formula still works, but be careful with signs in calculations.

Why This Matters

Geometric progressions appear everywhere: compound interest in finance, population growth in biology, radioactive decay in physics, and even in the design of algorithms (binary search halves the problem size each step — a GP with $r = 1/2$). Once you see the pattern of repeated multiplication, you'll spot GPs in many real-world contexts.

Geometric Progression is one of the two central sequence types in the NCERT Class 11 Mathematics chapter on Sequences and Series, and searches like "geometric progression: definition, formula and examples" or "GP sum of n terms important questions" point straight to this concept. It's also a regular fixture in JEE Main, CET, and other competitive exams, especially problems involving compound interest and infinite series.