Sum of the first n terms of an AP
Sum of the first n terms: Sₙ = n/2 [2a + (n − 1)d]. If you know the last term l, use Sₙ = n/2 (a + l).
How the formula is found (derivation)
This is the trick the young mathematician Gauss is said to have used. It can come as a 3-mark proof.
- 1Write the sum forwards: Sₙ = a + (a + d) + (a + 2d) + ... + [a + (n − 1)d]
- 2Write the same sum backwards: Sₙ = [a + (n − 1)d] + [a + (n − 2)d] + ... + a
- 3★Add the two lines term by term. Every pair adds up to the same value, 2a + (n − 1)d.first + last, second + second-last, and so on
- 4There are n such pairs, so 2Sₙ = n [2a + (n − 1)d].
- 5Divide by 2: Sₙ = n/2 [2a + (n − 1)d].
- 6Since a + (n − 1)d = l, the last term, this is also Sₙ = n/2 (a + l).
★ key step
Example 1: sum from a and d
Find the sum of the first 20 terms of the AP 7, 11, 15, ....
- a = 7, d = 11 − 7 = 4, n = 20
- Formula: Sₙ = n/2 [2a + (n − 1)d]
- S₂₀ = 20/2 [2 × 7 + 19 × 4]
- S₂₀ = 10 × [14 + 76] = 10 × 90
- S₂₀ = 900
S₂₀ = 900
Example 2: sum when the last term is known
For Van Mahotsav, groups of volunteers plant saplings. The first group plants 5, each group plants 3 more than the group before, and the last group plants 62. How many saplings are planted in all?
- The numbers 5, 8, 11, ..., 62 form an AP with a = 5, d = 3 and l = 62.
- First find n: 5 + (n − 1) × 3 = 62
- (n − 1) × 3 = 57, so n − 1 = 19 and n = 20
- Formula: Sₙ = n/2 (a + l)
- S₂₀ = 20/2 × (5 + 62) = 10 × 67 = 670
670 saplings are planted in all.
Example 3: how many terms give a sum?
How many terms of the AP 3, 7, 11, ... must be added to get a sum of 210?
- a = 3, d = 4, Sₙ = 210
- Sₙ = n/2 [2 × 3 + (n − 1) × 4] = n/2 (4n + 2) = n(2n + 1)
- So n(2n + 1) = 210, that is 2n² + n − 210 = 0
- Factorise: (n − 10)(2n + 21) = 0
- n = 10 or n = −21/2. A number of terms cannot be negative or a fraction, so n = 10.
- Check: S₁₀ = 10/2 × (6 + 36) = 5 × 42 = 210
10 terms
Useful results and traps
- aₙ = Sₙ − Sₙ₋₁: the nth term is the sum of n terms minus the sum of (n − 1) terms. Also a₁ = S₁.
- Sum of the first n natural numbers: n(n + 1)/2. So 1 + 2 + ... + 100 = 5050.
- If the equation gives two values of n, reject a negative or fractional value and say why.
- In a falling AP, two positive values of n can both be right: the extra terms add up to zero.
- Write Sₙ for a sum and aₙ for a term. Mixing them up is the most common mistake.
Save ₹ 5 on day 1, ₹ 10 on day 2, ₹ 15 on day 3 and so on. Sₙ tells you that you will have ₹ 2325 after 30 days without adding 30 numbers.
2 marks: “Find the sum of the first 12 multiples of 9.” AP 9, 18, ..., 108 with a = 9, d = 9, n = 12. S₁₂ = 12/2 × (9 + 108) = 6 × 117 = 702.
3 marks: “The sum of the first n terms of an AP is Sₙ = 3n² + 2n. Find its first term, common difference and 10th term.” a₁ = S₁ = 5. S₂ = 16, so a₂ = 16 − 5 = 11 and d = 6. a₁₀ = 5 + 9 × 6 = 59.
5 marks: “Derive Sₙ = n/2 [2a + (n − 1)d]. Use it to find the sum of all two-digit numbers that leave remainder 1 when divided by 4.” Write the sum forwards and backwards and add: n pairs of 2a + (n − 1)d. The numbers are 13, 17, ..., 97: a = 13, d = 4, n = 22. S₂₂ = 22/2 × (13 + 97) = 11 × 110 = 1210.
Forwards + backwards = n equal pairs. That one idea gives Sₙ = n/2 (first + last).