AP and GP Calculator

nth term (aₙ)
29
Sum of n terms (Sₙ)
155
Working
aₙ = a + (n − 1)d = 2 + (10 − 1) × 3 = 29
Sₙ = n/2 × [2a + (n − 1)d] = 10/2 × [2 × 2 + 9 × 3] = 155
First terms
2, 5, 8, 11, 14, 17, 20, 23, 26, 29

What are AP and GP?

An arithmetic progression (AP) adds the same number, the common difference d, to get each next term: 2, 5, 8, 11… A geometric progression (GP) multiplies by the same number, the common ratio r: 3, 6, 12, 24… This calculator finds the nth term and the sum of the first n terms of either, with the working as in Class 10 and 11 textbooks.

AP and GP formulas

AP: aₙ = a + (n − 1)d, Sₙ = n/2 × [2a + (n − 1)d] · GP: aₙ = arⁿ⁻¹, Sₙ = a(rⁿ − 1) ÷ (r − 1), S∞ = a ÷ (1 − r) when −1 < r < 1

a is the first term and n the number of terms. For an AP you can also use Sₙ = n/2 × (first term + last term).

Useful results

  • The sum of the first n natural numbers is n(n + 1)/2, so 1 + 2 + … + 100 = 5,050.
  • The sum of the first n odd numbers is n²: 1 + 3 + 5 + 7 = 16.
  • Three numbers in AP can be written a − d, a, a + d; three in GP as a/r, a, ar. This simplifies many textbook problems.
  • A GP with ratio between −1 and 1 shrinks towards 0, so its terms add up to a finite total: 1 + ½ + ¼ + … = 2.
  • Compound interest is a GP: each year's amount is the previous one × (1 + rate).

Worked example: the AP 2, 5, 8, …

  1. First term a = 2 and common difference d = 3
  2. 10th term: a₁₀ = 2 + (10 − 1) × 3 = 29
  3. Sum of 10 terms: S₁₀ = 10/2 × [2 × 2 + 9 × 3] = 5 × 31 = 155
  4. Check with Sₙ = n/2 × (first + last) = 5 × (2 + 29) = 155

Sums of common series

Standard formulas and their values for n = 10 and n = 100.

SeriesSum of n termsn = 10n = 100
Natural numbers 1, 2, 3, …n(n + 1)/2555,050
Odd numbers 1, 3, 5, …n²10010,000
Even numbers 2, 4, 6, …n(n + 1)11010,100
Squares 1², 2², 3², …n(n + 1)(2n + 1)/63853,38,350
Cubes 1³, 2³, 3³, …[n(n + 1)/2]²3,0252,55,02,500

The first three are APs; squares and cubes are not, but their sums are standard results.

Frequently Asked Questions

Use aₙ = a + (n − 1)d. For 2, 5, 8, …, the 10th term is 2 + 9 × 3 = 29.
Use Sₙ = n/2 × [2a + (n − 1)d], or n/2 × (first term + last term). The first 10 terms of 2, 5, 8, … add up to 155.
Use Sₙ = a(rⁿ − 1) ÷ (r − 1) when r isn't 1. For 3, 6, 12, … with 6 terms: 3 × (2⁶ − 1) ÷ 1 = 189.
Only when the common ratio is between −1 and 1, so the terms keep getting smaller. Then S∞ = a ÷ (1 − r): 1 + 0.5 + 0.25 + … = 2.
n(n + 1)/2 = 100 × 101 ÷ 2 = 5,050.