Geometric Progression (GP)

Master Geometric Progressions - nth term, sum formulas, infinite GP, and exponential growth applications

12 min read

Real-Life Hook: The Viral Video Phenomenon

You post a video online. On day 1, it gets 100 views. On day 2, it gets 300 views (3× more). On day 3, it gets 900 views (3× again). On day 4, 2700 views… This is Geometric Progression in action!

From compound interest that multiplies your wealth to bacteria doubling every hour, from radioactive decay to population growth, GP governs anything that grows or shrinks by a constant ratio. Understanding GP unlocks the mathematics of exponential change.

What is a Geometric Progression?

A Geometric Progression (GP) is a sequence where each term is obtained by multiplying the previous term by a constant value called the common ratio (r).

General Form: $a, ar, ar^2, ar^3, \ldots$

where:

  • $a$ = first term (non-zero)
  • $r$ = common ratio (non-zero)
  • $n$ = number of terms

Example: 2, 6, 18, 54, 162, … (here $a = 2$, $r = 3$)

Interactive Demo: Geometric Progression Explorer

GP Calculator & Visualizer

Core Formulas

1. nth Term Formula

$$\boxed{T_n = ar^{n-1}}$$

Memory Trick: “Always Raise to N-1” → $ar^{n-1}$

  • Start with $a$
  • Multiply by $r$ exactly $(n-1)$ times

2. Sum of n Terms (when $r \neq 1$)

$$\boxed{S_n = a\frac{r^n - 1}{r - 1} = a\frac{1 - r^n}{1 - r}}$$

Use which form?

  • If $r > 1$: Use $S_n = a\frac{r^n - 1}{r - 1}$ (positive denominator)
  • If $r < 1$: Use $S_n = a\frac{1 - r^n}{1 - r}$ (positive denominator)

When $r = 1$: All terms are equal to $a$, so $S_n = na$

Memory Trick: “Great Ratios Need Difference” → $\frac{r^n-1}{r-1}$

3. Common Ratio Test

$$\boxed{r = \frac{T_{n+1}}{T_n} = \frac{T_n}{T_{n-1}}}$$

A sequence is GP if the ratio of consecutive terms is constant.

Interactive Demo: Visualize Geometric Progression

Explore how GP terms grow or decay exponentially with different ratios.

Important Properties

1. Three Terms in GP

If $a, b, c$ are in GP, then:

$$\boxed{b^2 = ac}$$

or

$$b = \pm\sqrt{ac}$$

The middle term is the geometric mean of the outer terms.

2. Selection of Terms in GP

  • 3 terms: $\frac{a}{r}, a, ar$ (product = $a^3$)
  • 4 terms: $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ (product = $a^4$)
  • 5 terms: $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$ (product = $a^5$)

Advantage: Symmetric selection simplifies product calculations!

3. Product of Terms Equidistant from Ends

In a GP with $n$ terms:

$$T_k \cdot T_{n-k+1} = a \cdot l = \text{constant}$$

where $l$ is the last term.

Example: In GP with 10 terms, $T_2 \cdot T_9 = T_3 \cdot T_8 = T_1 \cdot T_{10}$

4. Geometric Mean (GM)

GM of two numbers $a$ and $b$:

$$\text{GM} = \sqrt{ab}$$

Insert $n$ GMs between $a$ and $b$:

Common ratio:

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

5. Important GP Relationships

  1. If each term of GP is multiplied/divided by same non-zero constant, result is still GP
  2. If each term is raised to same power, result is still GP
  3. Reciprocals of GP terms form a GP with ratio $\frac{1}{r}$
  4. Product of corresponding terms of two GPs is also a GP

Common Mistakes & How to Avoid Them

❌ Mistake 1: Power Confusion in nth Term

Wrong: $T_n = ar^n$ ✗

Correct: $T_n = ar^{n-1}$ ✓

Why? For $n=1$, we need $T_1 = a$, not $ar$!

Check: $T_1 = ar^{1-1} = ar^0 = a$ ✓

❌ Mistake 2: Sign Errors with Negative Ratio

Problem: Find 5th term of GP: 3, -6, 12, -24, …

Wrong: $r = \frac{-6}{3} = -2$, then $T_5 = 3(-2)^4 = 3 \times 16 = 48$ ✗

Correct: $r = -2$, so $T_5 = 3(-2)^4 = 3 \times 16 = 48$ ✓

Wait, that’s correct! Let me fix:

Wrong: $T_5 = 3(2)^4 = 48$ (forgetting negative sign) ✗

Correct: $r = -2$, so $T_5 = 3(-2)^{5-1} = 3(-2)^4 = 3 \times 16 = 48$ ✓

Note: $(-2)^4 = +16$ (even power → positive)

For $T_6 = 3(-2)^5 = 3 \times (-32) = -96$ (odd power → negative)

❌ Mistake 3: Wrong Sum Formula Selection

Problem: Sum first 5 terms of GP: 2, -6, 18, -54, …

Wrong: $r = -3$, using $S_n = a\frac{r^n-1}{r-1}$:

$S_5 = 2\frac{(-3)^5-1}{-3-1} = 2\frac{-243-1}{-4} = 2\frac{-244}{-4} = 2 \times 61 = 122$ ✓

Alternative (also correct): $S_5 = a\frac{1-r^n}{1-r} = 2\frac{1-(-3)^5}{1-(-3)} = 2\frac{1-(-243)}{4} = 2\frac{244}{4} = 122$ ✓

Tip: Both forms work! Choose the one that gives simpler arithmetic.

❌ Mistake 4: Forgetting r = 1 Case

Problem: Sum first 10 terms when $a = 5, r = 1$

Wrong: $S_{10} = 5\frac{1^{10}-1}{1-1} = \frac{0}{0}$ ✗ (undefined!)

Correct: When $r = 1$, all terms equal $a$, so $S_{10} = 10a = 10 \times 5 = 50$ ✓

Always check: Is $r = 1$? Use $S_n = na$ instead!

❌ Mistake 5: Sum of Infinite GP Conditions

Wrong: $S_\infty = \frac{a}{1-r}$ for any GP ✗

Correct: $S_\infty = \frac{a}{1-r}$ ONLY when $|r| < 1$ ✓

Counter-example: For $r = 2$, sum grows infinitely and doesn’t converge!

Problem-Solving Strategies

Strategy 1: Finding Unknown Terms

Problem: Three numbers in GP have sum 26 and product 216. Find the numbers.

Solution:

Let terms be $\frac{a}{r}, a, ar$ (smart selection!)

Product: $\frac{a}{r} \cdot a \cdot ar = a^3 = 216$ → $a = 6$

Sum: $\frac{6}{r} + 6 + 6r = 26$

$\frac{6}{r} + 6r = 20$

Multiply by $r$: $6 + 6r^2 = 20r$

$6r^2 - 20r + 6 = 0$

$3r^2 - 10r + 3 = 0$

$3r^2 - 9r - r + 3 = 0$

$3r(r-3) - 1(r-3) = 0$

$(3r-1)(r-3) = 0$

$r = 3$ or $r = \frac{1}{3}$

For $r = 3$: Terms are $\frac{6}{3}, 6, 18$ = $2, 6, 18$

For $r = \frac{1}{3}$: Terms are $18, 6, 2$

Answer: $2, 6, 18$ (or reverse order)

Strategy 2: Sum Problems with Powers

Problem: Find sum: $1 + 2 + 4 + 8 + \cdots + 2^{10}$

Solution:

This is GP with $a = 1, r = 2, n = 11$ (from $2^0$ to $2^{10}$, that’s 11 terms)

$S_{11} = 1 \cdot \frac{2^{11} - 1}{2-1} = 2^{11} - 1 = 2048 - 1 = 2047$

Shortcut: $1 + 2 + 4 + \cdots + 2^n = 2^{n+1} - 1$

Answer: 2047

Strategy 3: Finding Number of Terms

Problem: How many terms of GP $3, 3^2, 3^3, \ldots$ must be taken so that sum = 120?

Solution:

$a = 3, r = 3$

$S_n = 3 \cdot \frac{3^n - 1}{3-1} = 3 \cdot \frac{3^n-1}{2} = 120$

$\frac{3^n - 1}{2} = 40$

$3^n - 1 = 80$

$3^n = 81 = 3^4$

$n = 4$

Verification: $S_4 = 3 + 9 + 27 + 81 = 120$ ✓

Answer: 4 terms

Strategy 4: Mixed AP-GP Problems

Problem: Find $x$ if $1, x, 16$ are in GP and $x, 16, y$ are in AP.

Solution:

From GP condition: $x^2 = 1 \times 16 = 16$ → $x = \pm 4$

From AP condition: $2 \times 16 = x + y$ → $y = 32 - x$

If $x = 4$: $y = 32 - 4 = 28$

If $x = -4$: $y = 32 - (-4) = 36$

Answer: $(x, y) = (4, 28)$ or $(-4, 36)$

Practice Problems

Level 1: JEE Main Basics

Problem 1: Find the 8th term of the GP: 5, 15, 45, …

Solution

$a = 5$, $r = \frac{15}{5} = 3$

$T_8 = ar^{7} = 5 \times 3^7 = 5 \times 2187 = 10935$

Answer: 10935

Problem 2: Find the sum of first 6 terms of GP: 2, 6, 18, 54, …

Solution

$a = 2$, $r = 3$, $n = 6$

$S_6 = 2 \cdot \frac{3^6 - 1}{3 - 1} = 2 \cdot \frac{729 - 1}{2} = 728$

Answer: 728

Problem 3: If 5th term of GP is 48 and 8th term is 384, find the first term and common ratio.

Solution

$T_5 = ar^4 = 48$ … (1)

$T_8 = ar^7 = 384$ … (2)

Divide (2) by (1): $\frac{ar^7}{ar^4} = \frac{384}{48}$

$r^3 = 8$

$r = 2$

Substitute in (1): $a \times 2^4 = 48$

$16a = 48$

$a = 3$

Answer: $a = 3, r = 2$

Level 2: JEE Main/Advanced

Problem 4: The sum of first three terms of a GP is 13 and sum of their squares is 91. Find the GP.

Solution

Let GP be $\frac{a}{r}, a, ar$

Sum: $\frac{a}{r} + a + ar = 13$ … (1)

Sum of squares: $\frac{a^2}{r^2} + a^2 + a^2r^2 = 91$ … (2)

From (1): $a\left(\frac{1}{r} + 1 + r\right) = 13$

From (2): $a^2\left(\frac{1}{r^2} + 1 + r^2\right) = 91$

Note: $\frac{1}{r^2} + 1 + r^2 = \left(\frac{1}{r} + r\right)^2 - 2 + 1 = \left(\frac{1}{r} + r\right)^2 - 1$

Let $\frac{1}{r} + 1 + r = t$, then $\frac{1}{r} + r = t - 1$

From (1): $at = 13$

From (2): $a^2[(t-1)^2 - 1] = 91$

$a^2[t^2 - 2t + 1 - 1] = 91$

$a^2(t^2 - 2t) = 91$

$a^2 \cdot t(t-2) = 91$

$(at) \cdot a(t-2) = 91$

$13 \cdot a(t-2) = 91$

$a(t-2) = 7$

$at - 2a = 7$

$13 - 2a = 7$

$a = 3$

From $at = 13$: $t = \frac{13}{3}$

$\frac{1}{r} + 1 + r = \frac{13}{3}$

$\frac{1}{r} + r = \frac{10}{3}$

Multiply by $r$: $1 + r^2 = \frac{10r}{3}$

$3 + 3r^2 = 10r$

$3r^2 - 10r + 3 = 0$

$(3r-1)(r-3) = 0$

$r = 3$ or $r = \frac{1}{3}$

For $r = 3$: GP is $1, 3, 9$

For $r = \frac{1}{3}$: GP is $9, 3, 1$

Answer: $1, 3, 9$ or $9, 3, 1$

Problem 5: If $a, b, c$ are in GP and $\log_a x, \log_b x, \log_c x$ are in AP, find the common ratio of the GP.

Solution

Since $a, b, c$ are in GP: $b^2 = ac$ and $c = ar^2, b = ar$ for some $r$.

Given: $\log_a x, \log_b x, \log_c x$ are in AP

$2\log_b x = \log_a x + \log_c x$

Using change of base: $\log_b x = \frac{\ln x}{\ln b}$

$2 \cdot \frac{\ln x}{\ln b} = \frac{\ln x}{\ln a} + \frac{\ln x}{\ln c}$

Divide by $\ln x$ (assuming $x \neq 1$):

$\frac{2}{\ln b} = \frac{1}{\ln a} + \frac{1}{\ln c}$

$\frac{2}{\ln b} = \frac{\ln c + \ln a}{\ln a \cdot \ln c}$

$2\ln a \cdot \ln c = \ln b(\ln a + \ln c)$

$2\ln a \cdot \ln c = \ln b \cdot \ln(ac)$

Since $b^2 = ac$: $\ln b^2 = \ln(ac)$ → $2\ln b = \ln(ac)$

$2\ln a \cdot \ln c = \ln b \cdot 2\ln b$

$\ln a \cdot \ln c = (\ln b)^2$

Since $b = ar, c = ar^2$:

$\ln a \cdot \ln(ar^2) = (\ln ar)^2$

$\ln a \cdot (\ln a + 2\ln r) = (\ln a + \ln r)^2$

$(\ln a)^2 + 2\ln a \ln r = (\ln a)^2 + 2\ln a \ln r + (\ln r)^2$

$0 = (\ln r)^2$

$\ln r = 0$

$r = 1$

Wait, this means all terms are equal, which is trivial.

Let me reconsider. Actually, the condition “$a, b, c$ in GP” with the log condition might have a different solution. Let me try a different approach:

If $\frac{1}{\log_a x}, \frac{1}{\log_b x}, \frac{1}{\log_c x}$ (which are $\log_x a, \log_x b, \log_x c$) are in HP (reciprocals of AP are in HP), then:

$2\log_x b = \log_x a + \log_x c$

$\log_x b^2 = \log_x(ac)$

$b^2 = ac$ ✓ (which confirms GP condition!)

So actually, the answer is any common ratio $r$ works!

Answer: Any value of $r$ (the GP condition automatically satisfies the AP condition)

Actually, reviewing: the problem states they are in AP, not HP. Let me recalculate properly.

For AP: $2\log_b x = \log_a x + \log_c x$

This leads to $r = 1$ as shown above, which means the GP is constant.

Answer: $r = 1$

Level 3: JEE Advanced

Problem 6: Find the sum: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ to $\infty$

Solution

This is infinite GP with $a = 1, r = \frac{1}{2}$

Since $|r| = \frac{1}{2} < 1$, sum converges:

$S_\infty = \frac{a}{1-r} = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

Answer: 2

Problem 7: If $S_1, S_2, S_3, \ldots$ denote the sums of infinite geometric progressions with first terms $1, 2, 3, \ldots$ and common ratio $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ respectively, find $S_1 + S_2 + S_3 + \cdots + S_n$.

Solution

$S_k = \frac{k}{1 - \frac{1}{k+1}} = \frac{k}{\frac{k}{k+1}} = \frac{k(k+1)}{k} = k+1$

So: $S_1 = 2, S_2 = 3, S_3 = 4, \ldots, S_n = n+1$

Sum = $2 + 3 + 4 + \cdots + (n+1)$

This is sum of $(n+1-2+1) = n$ terms starting from 2.

$= \frac{n}{2}[2 + (n+1)] = \frac{n(n+3)}{2}$

Answer: $\frac{n(n+3)}{2}$

Problem 8: If the sum of first $n$ terms of a series is $2^n - 1$, show that it forms a GP and find the series.

Solution

$S_n = 2^n - 1$

$T_n = S_n - S_{n-1} = (2^n - 1) - (2^{n-1} - 1) = 2^n - 2^{n-1} = 2^{n-1}(2-1) = 2^{n-1}$

For $n = 1$: $T_1 = S_1 = 2^1 - 1 = 1$ ✓

Also, $T_1 = 2^{1-1} = 2^0 = 1$ ✓

The series is: $1, 2, 4, 8, 16, \ldots$ (GP with $a = 1, r = 2$)

Ratio: $\frac{T_{n+1}}{T_n} = \frac{2^n}{2^{n-1}} = 2$ (constant) ✓

Answer: GP with first term 1 and common ratio 2: $1, 2, 4, 8, 16, \ldots$

Cross-Topic Connections

GP problems often involve logarithms. If $a, b, c$ are in GP, then $\log a, \log b, \log c$ are in AP! See applications in logarithmic series.

Infinite GP convergence: $\lim_{n \to \infty} S_n = \frac{a}{1-r}$ when $|r| < 1$. This connects to Limits and Continuity.

Expanding $(1+x)^n$ and substituting special values creates GP patterns. See Binomial Theorem for connections.

Roots of unity form a GP on the complex plane. See Complex Numbers.

Quick Revision Checklist

  • Remember $T_n = ar^{n-1}$ (not $ar^n$!)
  • Know both sum forms: $\frac{r^n-1}{r-1}$ and $\frac{1-r^n}{1-r}$
  • Special case: When $r = 1$, $S_n = na$
  • Infinite GP: $S_\infty = \frac{a}{1-r}$ only when $|r| < 1$
  • Selection: 3 terms as $\frac{a}{r}, a, ar$ (product = $a^3$)
  • $b^2 = ac$ if $a, b, c$ in GP
  • Handle negative ratios carefully (check even/odd powers!)

Memory Palace Technique

Imagine a bacterial culture (GP = exponential growth):

  1. Petri Dish (First Term): You start with $a$ bacteria
  2. Doubling (Common Ratio): Every hour, bacteria multiply by $r$
  3. Time Machine (nth Term): After $n$ hours, you have $ar^{n-1}$ bacteria
  4. Total Count (Sum): To find total bacteria ever: $\frac{r^n-1}{r-1}$ formula
  5. Infinity (Infinite GP): If bacteria keep dying faster ($r < 1$), total approaches $\frac{a}{1-r}$

Final Tips for JEE

  1. Check $r = 1$ separately - sum formula doesn’t work!
  2. Infinite GP convergence - only when $|r| < 1$, never forget absolute value
  3. Sign management - negative $r$ alternates signs; watch even/odd powers
  4. Smart selection - use $\frac{a}{r}, a, ar$ for easier algebra with products
  5. Verify conditions - always check if your solution satisfies all given conditions
  6. Common ratio from two terms - $\frac{T_8}{T_5} = r^3$ (subtract indices!)
  7. Infinite decimals - $0.333\ldots = \frac{3}{9}$ is infinite GP application

Last Updated: October 18, 2025