If $p$th, $q$th, and $r$th terms of an A.P. and G.P. are both $a$, $b$ and $c$ respectively, show that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1$.

> [!TLDR] > When the $p$th, $q$th, and $r$th terms of an A.P. and a G.P. coincide at $a$, $b$, $c$ respectively, the relationship between positions and values forces $a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1$. The heart of this problem lies in exploiting the structural difference between arithmetic and geometric progressions. An A.P. grows by adding a constant difference, while a G.P. grows by multiplying a constant ratio. When three terms occupy the same positions in both progressions, the interplay between linear (A.P.) and exponential (G.P.) relationships creates a beautiful algebraic identity. The key insight: we'll express $a$, $b$, $c$ using both progression formulas, then eliminate the unknown parameters to reveal the hidden symmetry. --- ## Setting up the progressions 1. **Write the general terms** For an A.P. with first term $A$ and common difference $d$: $$T_n = A + (n-1)d$$ For a G.P. with first term $G$ and common ratio $r$: $$T_n = G \cdot r^{n-1}$$ 2. **Express the given conditions** Since the $p$th, $q$th, and $r$th terms are $a$, $b$, $c$ in both progressions: **A.P. conditions:** $$a = A + (p-1)d \quad \ldots (1)$$ $$b = A + (q-1)d \quad \ldots (2)$$ $$c = A + (r-1)d \quad \ldots (3)$$ **G.P. conditions:** $$a = G \cdot R^{p-1} \quad \ldots (4)$$ $$b = G \cdot R^{q-1} \quad \ldots (5)$$ $$c = G \cdot R^{r-1} \quad \ldots (6)$$ --- ## Extracting relationships from the A.P. 3. **Find differences between terms** Subtracting equation (1) from (2): $$b - a = (q - p)d$$ Subtracting (2) from (3): $$c - b = (r - q)d$$ Subtracting (1) from (3): $$c - a = (r - p)d$$ > [!NOTE] > These differences tell us how the values $a$, $b$, $c$ are spaced in terms of the position gaps $q-p$, $r-q$, and $r-p$. --- ## Extracting relationships from the G.P. 4. **Find ratios between terms** Dividing equation (5) by (4): $$\frac{b}{a} = R^{q-p}$$ Dividing (6) by (5): $$

If pth, qth, and rth terms of an A.P. and G.P. are – Class 11 Mathematics

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Concept understanding — Geometric Progression Arithmetic Progression

Comparing Arithmetic and Geometric Progressions

When you look at a sequence of numbers, the very first question to ask is: how does one term turn into the next? Two patterns cover almost every sequence you'll meet in Class 11 — adding a fixed number each time, or multiplying by a fixed number each time.

Two different rules of motion

An Arithmetic Progression (AP) moves by addition: each term is the previous term plus a constant common difference $d$ . If the first term is $a$ , the sequence is $a,\ a+d,\ a+2d,\ \dots$ , so the growth is a straight line — linear.
A Geometric Progression (GP) moves by multiplication: each term is the previous term times a constant common ratio $r$ . If the first term is $a$ , the sequence is $a,\ ar,\ ar^2,\ \dots$ , so the growth curves sharply upward (or shrinks toward zero) — exponential, not linear.

Note

This page is not the place to re-derive the individual formulas from scratch — the full nth-term and sum derivations for AP live under the Arithmetic Progression concept, and the equivalent derivations for GP live under the Geometric Progression concept. This page exists purely to help you tell the two patterns apart and choose the right one quickly.

The fastest way to tell them apart

Take any three consecutive terms $t_1, t_2, t_3$ of the sequence you're given:

Test for AP: compute $t_2-t_1$ and $t_3-t_2$ . If they're equal, you have an AP, and that equal value is $d$ .
Test for GP: compute $\dfrac{t_2}{t_1}$ and $\dfrac{t_3}{t_2}$ . If they're equal, you have a GP, and that equal value is $r$ .

If neither test succeeds, the sequence is neither an AP nor a GP on its own — it may still be built from a combination of the two (an arithmetico-geometric sequence), which needs its own separate technique.

Side-by-side

Feature	Arithmetic Progression	Geometric Progression
Rule to get the next term	Add $d$	Multiply by $r$
nth term	$a+(n-1)d$	$a\,r^{\,n-1}$
Sum of $n$ terms	$\dfrac{n}{2}[2a+(n-1)d]$	$a\,\dfrac{r^n-1}{r-1}$ (for $r\neq1$ )
Shape of growth	Straight line	Curve (exponential)
Allowed values of $d$ / $r$	Any real number, including $0$	Any real number except $0$

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When the $p$ th, $q$ th, and $r$ th terms of an A.P. and a G.P. coincide at $a$ , $b$ , $c$ respectively, the relationship between positions and values forces $a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1$ .

The heart of this problem lies in exploiting the structural difference between arithmetic and geometric progressions. An A.P. grows by adding a constant difference, while a G.P. grows by multiplying a constant ratio. When three terms occupy the same positions in both progressions, the interplay between linear (A.P.) and exponential (G.P.) relationships creates a beautiful algebraic identity.

The key insight: we'll express $a$ , $b$ , $c$ using both progression formulas, then eliminate the unknown parameters to reveal the hidden symmetry.

Setting up the progressions

Write the general terms

For an A.P. with first term $A$ and common difference $d$ :

T_n = A + (n-1)d

For a G.P. with first term $G$ and common ratio $r$ :

T_n = G \cdot r^{n-1}

Express the given conditions

Since the $p$ th, $q$ th, and $r$ th terms are $a$ , $b$ , $c$ in both progressions:

A.P. conditions:

a = A + (p-1)d \quad \ldots (1)

b = A + (q-1)d \quad \ldots (2)

c = A + (r-1)d \quad \ldots (3)

G.P. conditions:

a = G \cdot R^{p-1} \quad \ldots (4)

b = G \cdot R^{q-1} \quad \ldots (5)

c = G \cdot R^{r-1} \quad \ldots (6)

Extracting relationships from the A.P.

Find differences between terms

Subtracting equation (1) from (2):

b - a = (q - p)d

Subtracting (2) from (3):

c - b = (r - q)d

Subtracting (1) from (3):

c - a = (r - p)d

Note

These differences tell us how the values $a$ , $b$ , $c$ are spaced in terms of the position gaps $q-p$ , $r-q$ , and $r-p$ .

Extracting relationships from the G.P.

Find ratios between terms

Dividing equation (5) by (4):

\frac{b}{a} = R^{q-p}

Dividing (6) by (5):

\frac{c}{b} = R^{r-q}

Dividing (6) by (4):

\frac{c}{a} = R^{r-p}

Express the common ratio in terms of $a$ , $b$ , $c$

From the ratios above:

R^{q-p} = \frac{b}{a}, \quad R^{r-q} = \frac{c}{b}, \quad R^{r-p} = \frac{c}{a}

Constructing the target expression

Rewrite $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}$ using logarithms

Taking logarithm of the expression we need to prove equals 1:

\log(a^{b-c} \cdot b^{c-a} \cdot c^{a-b}) = (b-c)\log a + (c-a)\log b + (a-b)\log c

Substitute the G.P. relationships

From step 4, we can write:

\log a = \log G + (p-1)\log R

\log b = \log G + (q-1)\log R

\log c = \log G + (r-1)\log R

Expand the logarithmic expression …

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Q.If pppth, qqqth, and rrrth terms of an A.P. and G.P. are both aaa, bbb and ccc respectively, show that ab−c⋅bc−a⋅ca−b=1a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1ab−c⋅bc−a⋅ca−b=1.

Setting up the progressions

Extracting relationships from the A.P.

Extracting relationships from the G.P.

Constructing the target expression

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Q.If $p$ th, $q$ th, and $r$ th terms of an A.P. and G.P. are both $a$ , $b$ and $c$ respectively, show that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1$ .

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