Geometric Progression: Formulas, nth Term, and Solved Examples

A geometric progression is a sequence where each term equals the previous term multiplied by a fixed constant called the common ratio. Every GP formula used in placement aptitude tests builds from that definition.

What Is a Geometric Progression?

A sequence is a geometric progression (GP) when the ratio between any pair of consecutive terms is constant. That constant ratio is called the common ratio, denoted r.

The general form of a GP with first term a and common ratio r is:

a, ar, ar^2, ar^3, ar^4, …

Two straightforward examples:

• 2, 10, 50, 250, … — each term is 5 times the previous, so r = 5.
• 20, 10, 5, 2.5, 1.25, … — each term is half the previous, so r = 1/2.

To confirm a sequence is a GP, divide any term by the one before it. If the ratio is the same for every consecutive pair, the sequence is a GP.

Geometric progressions appear in the NCERT Class 11 Mathematics, Chapter 9 curriculum on Sequences and Series and on every major campus placement aptitude test.

Geometric Progression Formulas

Three formulas cover every standard GP problem on placement tests.

nth Term of a GP

The nth term (Tn) of a GP with first term a and common ratio r:

Tn = a × r^(n-1)

The exponent is (n-1), not n, because the first term (n=1) has r^0 = 1 as its multiplier.

Sum of the First n Terms

For r not equal to 1, the sum Sn of the first n terms is:

Sn = a × (r^n - 1) / (r - 1)

An equivalent form when r is less than 1: Sn = a × (1 - r^n) / (1 - r). Both are algebraically identical; the second form avoids negative numerators when r is a fraction.

For r = 1 (every term equals a):

Sn = n × a

Sum of an Infinite GP

When the absolute value of r is less than 1, the series converges to a finite sum:

S∞ = a / (1 - r)

When |r| is 1 or greater, the series diverges. The formula is undefined for those cases.

Khan Academy’s geometric sequences reference covers step-by-step derivations of these formulas for students who prefer worked derivations alongside the results.

Properties of the Common Ratio

How a GP behaves depends entirely on r:

• r greater than 1: the sequence grows exponentially (bacterial populations, compound interest at positive rates).
• 0 less than r less than 1: the sequence decays toward zero (bouncing ball losing height, drug concentration falling between doses).
• r equals 1: all terms are identical; the sum grows in direct proportion to n.
• r less than 0: terms alternate between positive and negative values.
• r equals -1: terms alternate between +a and -a with no growth or decay; partial sums oscillate between a and 0.

Placement test questions frequently ask which of four options correctly describes a given GP’s behaviour. Knowing these five cases by their r-ranges answers those questions directly.

Geometric Mean and the GP–AP Bridge

The geometric mean (GM) of two positive numbers a and c is:

GM = square root of (a × c)

If three numbers a, b, c are in GP, then b divided by a equals c divided by b, which gives b squared = a times c. This condition is the standard test for whether three numbers form a GP.

To insert k geometric means between two numbers p and q, set up a GP with (k + 2) terms where the first term is p and the last is q. Solve for r using r^(k+1) = q/p, then generate the intermediate terms.

The logarithm connection: if a, ar, ar^2 are in GP, their logarithms log(a), log(a) + log(r), log(a) + 2log(r) form an arithmetic progression with common difference log(r). This GP-AP duality appears in logarithm-based aptitude questions.

Placement preparation requires fluency in both aptitude topics. The Time and Work aptitude guide on FACE Prep covers the rate-addition method that pairs with GP in many multi-topic placement tests.

Worked Examples

Finding Terms

• Example 1. Sequence: 3, 6, 12, 24. Is it a GP? Find the common ratio.

  • Check ratio: 6/3 = 2, 12/6 = 2, 24/12 = 2. Ratio is constant.
  • Yes, it is a GP with r = 2.

• Example 2. For the GP 3, 6, 12, 24, …, find the 10th term.

  • a = 3, r = 2, n = 10
  • T10 = 3 × 2^(10-1) = 3 × 2^9 = 3 × 512 = 1,536

• Example 3. The 3rd term of a GP is 18 and the 6th term is 486. Find the first term and common ratio.

  • T3 = ar^2 = 18; T6 = ar^5 = 486
  • Divide: T6/T3 = r^3 = 486/18 = 27, so r = 3
  • From T3: a × 9 = 18, so a = 2
  • The GP is 2, 6, 18, 54, 162, 486, …

Sums of Finite GP

• Example 4. Find the sum of the first 6 terms of 2, 6, 18, 54, …

  • a = 2, r = 3, n = 6
  • S6 = 2 × (3^6 - 1) / (3 - 1) = 2 × (729 - 1) / 2 = 729 - 1 = 728

• Example 5. A sequence has 8 terms each equal to 5. Find the sum.

  • r = 1, so use Sn = n × a
  • S8 = 8 × 5 = 40

Infinite Series

• Example 6. Find the sum of the infinite series 12, 4, 4/3, 4/9, …

  • r = 4/12 = 1/3. Since |r| = 1/3 is less than 1, the series converges.
  • S∞ = 12 / (1 - 1/3) = 12 / (2/3) = 18

• Example 7. The sum of an infinite GP is 48 and the sum of its first two terms is 36. Find the second term.

  • a / (1 - r) = 48, so a = 48(1 - r)
  • a + ar = 36, so a(1 + r) = 36
  • Substituting: 48(1 - r)(1 + r) = 36, which gives 48(1 - r^2) = 36
  • 1 - r^2 = 36/48 = 3/4, so r^2 = 1/4, giving r = 1/2
  • a = 48 × (1 - 1/2) = 24
  • Second term = ar = 24 × 1/2 = 12

Geometric Mean Problems

• Example 8. Find the geometric mean of 4 and 9.

  • GM = square root of (4 × 9) = square root of 36 = 6
  • Verification: 4, 6, 9 has ratio 6/4 = 9/6 = 1.5. Confirmed as a GP.

• Example 9. Insert three geometric means between 2 and 81/8.

  • Sequence with inserted means: 2, x1, x2, x3, 81/8
  • This is a 5-term GP: a = 2, T5 = ar^4 = 81/8
  • r^4 = (81/8) / 2 = 81/16. Since 3^4 = 81 and 2^4 = 16, r = 3/2
  • x1 = 2 × 3/2 = 3
  • x2 = 2 × (3/2)^2 = 2 × 9/4 = 9/2
  • x3 = 2 × (3/2)^3 = 2 × 27/8 = 27/4
  • The three geometric means are 3, 9/2, and 27/4

Real-World Applications

• Example 10. A bacterial culture starts with 50 bacteria and doubles every hour. How many bacteria are present after 12 hours?

  • a = 50, r = 2, n = 12
  • T12 = 50 × 2^(12-1) = 50 × 2^11 = 50 × 2,048 = 102,400 bacteria

• Example 11. A principal of Rs. 10,000 is invested at 10% per annum compound interest. Find the amount after 3 years.

  • Each year, the amount is multiplied by (1 + 10/100) = 1.1, forming a GP with r = 1.1.
  • Amount after 3 years = 10,000 × (1.1)^3 = 10,000 × 1.331 = Rs. 13,310

• Example 12. A ball is dropped from 128 metres. After each bounce it rises to half the height of the previous fall. Find the total distance travelled.

  • Initial fall: 128 m.
  • Subsequent bounces: rise and fall of 64 m, then 32 m, then 16 m, forming an infinite GP with first term 64 and r = 1/2.
  • Total bounce distance (up + down): 2 × [64 / (1 - 1/2)] = 2 × 128 = 256 m
  • Total distance = 128 + 256 = 384 metres

The Campus Placement Evaluation Test from FACE Prep benchmarks your current aptitude level across topics including GP, before your placement preparation cycle starts.

The compound interest in Example 11 and the convergent bounce sum in Example 12 follow the same mathematical structure: a base value scaled by a fixed ratio over discrete steps. That structure appears in finance, population modelling, and in the probability calculations behind language model token selection. TinkerLLM (Rs. 299 on launch) is a live Python environment where you can run these patterns in code, watching how the numbers change as r approaches 1 from below or as n grows large. Faster feedback than working through them on paper.