For $a, b, c$ to be in G.P. the value of $\dfrac{a-b}{b-c}$ is equal to .............. .

For a, b, c to be in G.P. the value of (a-b)/(b-c) is equal to .............. .

NCERT Exemplar · Q27

a, b, c

Sikkim CbseShort· 2mImportance★★★★★est

92% · 105/114 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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

Start your 14-day free trial to unlock the full solution →

When three terms $a, b, c$ are in Geometric Progression (G.P.), their defining property is that the square of the middle term equals the product of the other two ( $b^2 = ac$ ). Using this property, the expression $\frac{a-b}{b-c}$ simplifies to $\boxed{\frac{a}{b}}$ .

Let's understand what it means for three numbers to be in a Geometric Progression (G.P.) before we tackle the expression.

A sequence of numbers is said to be in G.P. if the ratio of any term to its preceding term is constant. This constant ratio is called the common ratio, usually denoted by $r$ .

For three terms $a, b, c$ to be in G.P., this means:

The ratio of the second term to the first term is equal to the ratio of the third term to the second term.

Mathematically, this is expressed as:

\frac{b}{a} = \frac{c}{b} = r

where $r$ is the common ratio.

This fundamental property gives us the key relationship for terms in G.P.:

\frac{b}{a} = \frac{c}{b}

Cross-multiplying these terms, we get:

b \cdot b = a \cdot c

b^2 = ac

This is the defining characteristic for three terms $a, b, c$ to be in G.P. We will use this property to simplify the given expression.

Now, let's work through the problem step-by-step.

Identify the G.P. property:

Since $a, b, c$ are in G.P., we know that $b^2 = ac$ . This is the most direct relationship we can use.
Express one variable in terms of others:

From $b^2 = ac$ , we can express $c$ in terms of $a$ and $b$ :

c = \frac{b^2}{a}

This substitution will help us simplify the denominator of the given expression.

3. Substitute into the given expression:

The expression we need to evaluate is $\dfrac{a-b}{b-c}$ .

Substitute $c = \frac{b^2}{a}$ into the denominator:

\frac{a-b}{b-c} = \frac{a-b}{b - \frac{b^2}{a}}

Simplify the denominator: To simplify the denominator, find a common denominator for $b - \frac{b^2}{a}$ :

b - \frac{b^2}{a} = \frac{b \cdot a}{a} - \frac{b^2}{a} = \frac{ab - b^2}{a}

Now, substitute this back into the main expression:

\frac{a-b}{b - \frac{b^2}{a}} = \frac{a-b}{\frac{ab - b^2}{a}}

Perform division and factor: Dividing by a fraction is equivalent to multiplying by its reciprocal:

\frac{a-b}{\frac{ab - b^2}{a}} = (a-b) \cdot \frac{a}{ab - b^2}

Now, factor out $b$ from the denominator $ab - b^2$:

ab - b^2 = b(a-b)

Substitute this back:

(a-b) \cdot \frac{a}{b(a-b)}

Cancel common terms: …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ26 — The lengths of three unequal edges of a rectangular solid block are in…→Next questionQ28 — The sum of terms equidistant from the beginning and end in an A.P. is …

Q.For a,b,ca, b, ca,b,c to be in G.P. the value of a−bb−c\dfrac{a-b}{b-c}b−ca−b​ is equal to .............. .

🎓Unlock everything free for 14 days

Q.For $a, b, c$ to be in G.P. the value of $\dfrac{a-b}{b-c}$ is equal to .............. .

Unlock everything free for 14 days