Geometric progressions model compound growth, population and decay, and are a Class 11 topic. Enter the first term, common ratio and number of terms to get the nth term, the sum, and the infinite sum when it exists.
Enter the first term and the common ratio.
Enter the number of terms.
Read the nth term, the sum and the infinite sum.
nth term: aₙ = a × r^(n − 1)
Sum: Sₙ = a(rⁿ − 1) ÷ (r − 1)
Infinite sum: S∞ = a ÷ (1 − r), when |r| < 1
In a geometric progression, or GP, each term is the previous one multiplied by the same number, the common ratio r. 3, 6, 12, 24 has r = 2, and 800, 400, 200, 100 has r = 0.5. This calculator takes the first term a, the ratio r and the number of terms n, and gives the nth term, the sum of n terms and, when −1 < r < 1, the sum to infinity.
GPs are a Class 11 topic and the mathematics of anything that grows or shrinks by a percentage: compound interest, inflation, population growth, depreciation of a vehicle's value, the fading bounce of a ball, radioactive decay and viral spread. Understanding them explains why small repeated percentages produce such large differences over time. The calculator also handles r = 1, where every term is the same, and flags when an infinite sum does not exist.
1. Find the first term a and the common ratio r = (second term ÷ first term). Check the ratio is the same throughout.
2. Find the nth term: aₙ = a × r^(n − 1).
3. Find the sum of n terms: Sₙ = a(rⁿ − 1) ÷ (r − 1), valid when r ≠ 1.
4. If r = 1, every term equals a and Sₙ = n × a.
5. If −1 < r < 1, the sum to infinity is S∞ = a ÷ (1 − r).
6. If |r| ≥ 1 (and a is not zero), the terms do not shrink and there is no finite infinite sum.
Write S = a + ar + ar² + … + ar^(n−1). Multiply every term by r to get rS = ar + ar² + … + arⁿ. The two lists share every term except the first of S and the last of rS. Subtracting, rS − S = arⁿ − a, so S(r − 1) = a(rⁿ − 1) and S = a(rⁿ − 1) ÷ (r − 1). The equivalent form a(1 − rⁿ) ÷ (1 − r) is neater when r is less than 1. Both fail at r = 1, which is handled separately.
If −1 < r < 1, rⁿ gets closer and closer to zero as n grows, so the sum formula approaches a(0 − 1) ÷ (r − 1) = a ÷ (1 − r). The terms shrink fast enough for the total to settle at a finite value. Walking halfway to a wall, then half the remaining distance and so on, covers 1 + ½ + ¼ + … = 2 times the first step. If |r| ≥ 1 the terms stay large or keep growing, so the total runs off without limit or keeps oscillating.
Money invested at 10% a year compounds as P, 1.1P, 1.21P and so on, a GP with r = 1.1. That is why the future value formula P(1 + i)ⁿ has the same shape as the nth term. Regular equal deposits into such an account form a GP sum, which is how recurring deposit and SIP maturity formulas are derived. A negative ratio makes the terms alternate in sign, for example 5, −10, 20, −40.
Aarav walks 1,000 m towards a gate on the first leg, and on every later leg he walks half the distance of the leg before; his teacher asks for the total after 6 legs and in the long run.
nth term: aₙ = a × r^(n − 1) a6 = 1000 × 0.5^5 = 31.25
Sum of n terms: Sₙ = a(rⁿ − 1) ÷ (r − 1) S6 = 1000 × (0.5^6 − 1) ÷ (0.5 − 1) = 1,968.75
Sum to infinity: S∞ = a ÷ (1 − r) = 1000 ÷ 0.5 = 2,000
Answer: Sum of first 6 terms 1,968.75; 6th term 31.25; Sum to infinity 2,000
Using rⁿ instead of r^(n − 1) for the nth term.
Finding r by subtraction, which is the AP method, instead of by division.
Applying the infinite-sum formula when |r| ≥ 1, which gives a meaningless answer.
Entering a growth rate of 8% as r = 8 or 0.08 instead of 1.08.
Mixing up the nth term with the sum when a question asks how much in total.
Modelling compound interest, inflation and investment growth year by year.
Estimating depreciation of a car or machine that loses a fixed percentage each year.
Working out total distance travelled by a bouncing ball or any halving process.
Modelling population growth, bacterial doubling and radioactive decay.
Deriving recurring deposit and loan instalment formulas from GP sums.
When does an infinite GP have a sum?
Only when the ratio lies strictly between −1 and 1.
How do I find r?
Divide any term by the previous one: r = a₂ ÷ a₁.