Geometric progression

What is a geometric progression?

In plain words: in a G.P. you get each term by multiplying the one before it by the same number r; in an A.P. you add the same number instead. NCERT’s three examples (page 139) are 2, 4, 8, 16, … with r = 2; 1/9, −1/27, 1/81, −1/243, … with r = −1/3; and 0.01, 0.0001, 0.000001, … with r = 0.01.

It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it.

NCERT Mathematics (Class 11), Section 8.4, page 139

A negative r makes the signs alternate, and a ratio between −1 and 1 makes the terms shrink in size towards 0. The numbers of subsets of sets with 0, 1, 2, 3, … elements, 1, 2, 4, 8, …, form a G.P. with r = 2 (see the power set).

The general term

Each step multiplies by r, so the nth term is a multiplied by r exactly n − 1 times (NCERT Section 8.4.1, page 140): a₁ = a, a₂ = ar, a₃ = ar², …, a₁₆ = ar¹⁵.

NCERT Example 4 (page 140): in 5, 25, 125, …, a = 5 and r = 5, so a₁₀ = 5 × 5⁹ = 5¹⁰ and a_n = 5 × 5ⁿ⁻¹ = 5ⁿ.

NCERT Example 5 (page 141): in 2, 8, 32, …, a = 2 and r = 4. If 131 072 is the nth term, 2 × 4ⁿ⁻¹ = 131 072, so 4ⁿ⁻¹ = 65 536 = 4⁸, n − 1 = 8 and n = 9: 131 072 is the 9th term.

Sum to n terms

Let S_n = a + ar + ar² + … + arⁿ⁻¹ (NCERT Section 8.4.2, page 140). If r = 1, every term is a and S_n = na. If r ≠ 1, multiply by r: rS_n = ar + ar² + … + arⁿ. The two sums share every term except a and arⁿ, so subtracting gives (1 − r)S_n = a − arⁿ = a(1 − rⁿ).

Use the first form when r < 1 and the second when r > 1, so the denominator is positive; both give the same number. NCERT page 140 prints only the second form: after “This gives” it leaves a blank where S_n = a(1 − rⁿ)/(1 − r) belongs, and then prints “or” followed by S_n = a(rⁿ − 1)/(r − 1). Both forms are in NCERT’s Summary (page 149).

NCERT Example 11 (page 143), using the idealized assumption that no ancestor occupies multiple positions in the family tree: a person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Over the ten generations before his own, a = 2, r = 2, n = 10, so S₁₀ = 2(2¹⁰ − 1) = 2046 ancestors.

Sum to infinity (JEE). If −1 < r < 1, the powers rⁿ get as close to 0 as we like as n grows, so S_n = a(1 − rⁿ)/(1 − r) gets as close as we like to a/(1 − r), the sum of the infinite G.P. The 2026–27 NCERT chapter does not cover it (the book gives it only in Appendix 1, A.1.3, page 316). It turns a recurring decimal into a fraction: 0.333… = 0.3 + 0.03 + 0.003 + … = 0.3/(1 − 0.1) = 1/3. If r ≥ 1 or r ≤ −1 the infinite G.P. has no sum.

Geometric mean

The geometric mean (G.M.) of two positive numbers a and b is √(ab) (NCERT Section 8.4.3, page 143). The G.M. of 2 and 8 is √16 = 4, and 2, 4, 8 are consecutive terms of a G.P.: a, G, b form a G.P. exactly when G² = ab. The G.M. is never more than the A.M. (a + b)/2.

Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

NCERT Mathematics (Class 11), Section 8.4.3, page 143

To insert n numbers G₁, G₂, …, G_n between a and b, note that b is then the (n + 2)th term: b = arⁿ⁺¹. For positive a and b the G.M.s are taken positive, so r = (b/a)^1/(n+1). When n + 1 is even, r = −(b/a)^1/(n+1) also gives a G.P., with alternating signs, as in Example 6. The product of the n G.M.s is (√(ab))ⁿ, because terms the same distance from the two ends multiply to ab.

Solved examples

Common mistakes to avoid

Quick revision summary

• A G.P. has non-zero terms with a constant ratio r = a_k+1/a_k; it is a, ar, ar², … and its nth term is arⁿ⁻¹.
• S_n = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1) for r ≠ 1, and na for r = 1.
• The G.M. of positive a and b is √(ab); n G.M.s between a and b use r = (b/a)^1/(n+1).

Questions students ask

What is the difference between an arithmetic progression and a geometric progression?

An arithmetic progression adds the same common difference d to get each next term, while a geometric progression multiplies by the same common ratio r: 4, 7, 10, 13, … is an A.P. with d = 3, and 4, 12, 36, 108, … is a G.P. with r = 3. So the nth term is a + (n − 1)d in an A.P. but arⁿ⁻¹ in a G.P., and a sequence is tested by checking for equal differences in one case and equal ratios in the other. An A.P. changes by equal steps, while a G.P. of positive terms with r > 1 grows faster and faster. A constant sequence of non-zero terms, such as 5, 5, 5, …, is both, with d = 0 and r = 1.

Why is a geometric progression called geometric?

The name geometric progression goes back to the geometric mean, one of the three Pythagorean means (arithmetic, geometric and harmonic): in a G.P. of positive terms, each term with a term on each side is the geometric mean of those two, as each such term of an A.P. is their arithmetic mean. In 2, 10, 50, 250, …, 10 = √(2 × 50) and 50 = √(10 × 250). Why the mean itself is called geometric is less certain, but √(ab) has a geometric meaning: it is the side of a square with the same area as an a by b rectangle, which Euclid calls the mean proportional. With a negative ratio the signs alternate, but a_k² = a_k−1a_k+1 still holds.

What is the sum of an infinite G.P., and when does it exist?

The sum of an infinite G.P. a + ar + ar² + … is a/(1 − r), and it exists only when −1 < r < 1, because only then do the powers rⁿ shrink towards 0 and S_n = a(1 − rⁿ)/(1 − r) settle on one number. For 12 − 6 + 3 − 3/2 + …, with r = −1/2, the sum is 12/(1 + 1/2) = 8, and the partial sums 12, 6, 9, 7.5, … close in on 8. If r = 1 the partial sums na grow without limit in size, if r = −1 they jump between a and 0, and if |r| > 1 the terms grow, so there is no sum. NCERT (2026–27) gives this sum only in Appendix 1 (A.1.3); the JEE Main syllabus (Unit 6) names geometric progressions without listing this formula separately.

How do you find the sum of a G.P. when its last term is given?

When the last term l of a G.P. is known, its sum is S_n = (lr − a)/(r − 1) for r ≠ 1, so the number of terms need not be found first. This follows from S_n = a(rⁿ − 1)/(r − 1), because l = arⁿ⁻¹ gives arⁿ = lr. For 2 + 6 + 18 + … + 1458, with r = 3, the sum is (1458 × 3 − 2)/(3 − 1) = 4372/2 = 2186. When r < 1 the same formula is easier to use as (a − lr)/(1 − r).

Is a geometric progression the same as exponential growth?

A geometric progression with a positive first term and positive common ratio models exponential growth (r > 1) or decay (0 < r < 1) counted in whole steps; r = 1 gives a constant sequence: its nth term arⁿ⁻¹ is an exponential function of n, taken only at n = 1, 2, 3 and so on. Anything that is multiplied by the same factor in each equal step of time forms a G.P., such as money at 8% a year compounded annually, multiplied by r = 1.08 each year, or a sample whose mass halves every hour, with r = 1/2. The difference is that an exponential function such as 2^x is defined for every real x, while a G.P. is a sequence, a function on the natural numbers only.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the 12th term of a G.P. whose 8th term is 192 and common ratio is 2 (NCERT Exercise 8.2, Q2).

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   a₁₂ = a₈ × r⁴ = 192 × 16 = 3072.

2. Q2. Which term of (a) 2, 2√2, 4, … is 128; (b) √3, 3, 3√3, … is 729; (c) 1/3, 1/9, 1/27, … is 1/19 683 (NCERT Exercise 8.2, Q5)?

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   (a) r = √2: 2(√2)ⁿ⁻¹ = 128 gives (√2)ⁿ⁻¹ = 64 = (√2)¹², so n = 13. (b) r = √3: (√3)ⁿ = 729 = (√3)¹², so n = 12. (c) r = 1/3: (1/3)ⁿ = 1/3⁹, so n = 9.

3. Q3. How many terms of the G.P. 3, 3², 3³, … are needed to give the sum 120 (NCERT Exercise 8.2, Q13)?

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   S_n = 3(3ⁿ − 1)/2 = 120 gives 3ⁿ = 81, so n = 4 (3 + 9 + 27 + 81 = 120).

4. Q4. Given a G.P. with a = 729 and 7th term 64, determine S₇ (NCERT Exercise 8.2, Q15).

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   729r⁶ = 64, so r = 2/3 or −2/3.

   With r = 2/3,

   S₇ = 729[1 − (2/3)⁷]/(1/3)

   = 2187 − 128 = 2059.

   With r = −2/3,

   S₇ = 729[1 + (2/3)⁷]/(5/3)

   = 463.