If $a$, $b$, $c$, $d$ are in G.P, prove that $(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$ are in G.P.

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Accepted Answer

> [!TLDR]
> If $a, b, c, d$ are in GP, then each term is the previous multiplied by a constant ratio $r$. Raising to the $n$th power preserves the geometric structure, so the three given expressions also form a GP with common ratio $r^n$.

## Why This Works — The Core Idea

A geometric progression is defined by a constant ratio between consecutive terms. If $a, b, c, d$ are in GP, then:

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

where $r$ is the common ratio. The key insight: when you raise each term to the $n$th power, the ratio between consecutive powers becomes $r^n$, not $r$. So $a^n, b^n, c^n, d^n$ are also in GP, with common ratio $r^n$.

Now look at the expressions we need to check: $(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$. Each is a sum of two consecutive terms from the GP $a^n, b^n, c^n, d^n$. The question is: does adding consecutive terms of a GP produce another GP? Not always — but here it does, because of the specific structure.

Let's see why.

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## Step-by-Step Proof

**1. Set up the GP with a common ratio**

Since $a, b, c, d$ are in GP, let the common ratio be $r$. Then:

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

This is the standard parametrisation. It captures the entire progression in terms of the first term $a$ and the ratio $r$.

**2. Write the three expressions in terms of $a$ and $r$**

First expression: $a^n + b^n = a^n + (ar)^n = a^n + a^n r^n = a^n(1 + r^n)$

Second expression: $b^n + c^n = (ar)^n + (ar^2)^n = a^n r^n + a^n r^{2n} = a^n r^n (1 + r^n)$

Third expression: $c^n + d^n = (ar^2)^n + (ar^3)^n = a^n r^{2n} + a^n r^{3n} = a^n r^{2n} (1 + r^n)$

Notice the common factor $a^n(1 + r^n)$ appears in all three — that's the key.

**3. Check the ratio between consecutive expressions**

For three quantities to be in GP, the ratio of the second to the first must equal the ratio of the third to the second.

Compute the first ratio:

$$\frac{b^n + c^n}{a^n + b^n} = \frac{a^n r^n (1 + r^n)}{a^n (1 + r^n)} = r^n$$

Compute the second ra

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

Q.If $a$ , $b$ , $c$ , $d$ are in G.P, prove that $(a^n + b^n)$ , $(b^n + c^n)$ , $(c^n + d^n)$ are in G.P.

Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Precise Definition

The $n$ -th Term

Sum of $n$ Terms

Sum of an Infinite GP

Quick Reference Table

Why This Works — The Core Idea

Step-by-Step Proof

Q.If aaa, bbb, ccc, ddd are in G.P, prove that (an+bn)(a^n + b^n)(an+bn), (bn+cn)(b^n + c^n)(bn+cn), (cn+dn)(c^n + d^n)(cn+dn) are in G.P.

Why This Works — The Core Idea

Step-by-Step Proof

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Q.If $a$ , $b$ , $c$ , $d$ are in G.P, prove that $(a^n + b^n)$ , $(b^n + c^n)$ , $(c^n + d^n)$ are in G.P.

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