If $a$, $b$, $c$ and $d$ are in G.P. show that $(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$.

If a, b, c and d are in G.P. show that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Exercise 8.2 · Q25

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Concept understanding — Geometric Progression

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:

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

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

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

T_n = a \cdot r^{\,n-1}

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.

n

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:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

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

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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Using the common ratio property of a G.P., we express all terms in terms of $a$ and $r$ , then expand both sides to show they are identical — proving the given identity.

Why This Works

When four numbers are in Geometric Progression, they follow a simple multiplicative pattern: each term is the previous term multiplied by a fixed ratio. This structure lets us rewrite every term using just the first term $a$ and the common ratio $r$ . Once we do that, the messy-looking expression becomes a clean algebraic identity — no guesswork, just straightforward expansion and cancellation.

The key insight: a G.P. is defined by a single ratio, so all terms are powers of $r$ times $a$ . That’s the only fact we need.

Step-by-Step Proof

1. Set up the G.P. notation

Let the common ratio be $r$ . Then:

b = ar, \quad c = ar^2, \quad d = ar^3

2. Write each bracket in terms of $a$ and $r$

First bracket:

a^2 + b^2 + c^2 = a^2 + (ar)^2 + (ar^2)^2 = a^2 + a^2 r^2 + a^2 r^4 = a^2(1 + r^2 + r^4)

Second bracket:

b^2 + c^2 + d^2 = (ar)^2 + (ar^2)^2 + (ar^3)^2 = a^2 r^2 + a^2 r^4 + a^2 r^6 = a^2 r^2 (1 + r^2 + r^4)

Right-hand side:

ab + bc + cd = a(ar) + (ar)(ar^2) + (ar^2)(ar^3) = a^2 r + a^2 r^3 + a^2 r^5 = a^2 r (1 + r^2 + r^4)

Tip

Notice the common factor $(1 + r^2 + r^4)$ appears in all three expressions. This is the pattern that makes the identity work — the G.P. structure creates a repeating polynomial in $r^2$ .

3. Compute the left-hand side (LHS) …

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Q.If aaa, bbb, ccc and ddd are in G.P. show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2(a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2.

Why This Works

Step-by-Step Proof

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Q.If $a$ , $b$ , $c$ and $d$ are in G.P. show that $(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$ .

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