8 min readConcept sheetPart II (2 marks); word problems and 'how many terms' questions are Part III (5 marks)

Sum of an arithmetic series

Sum of the first n terms of an AP: Sₙ = (n/2)[2a + (n − 1)d], or Sₙ = (n/2)(a + l) if the last term l is known.

Sum formulaSₙ = (n/2)[2a + (n − 1)d]

With the last termSₙ = (n/2)(a + l)

nth term from sumstₙ = Sₙ − Sₙ₋₁, and t₁ = S₁

SeriesThe sum of the terms of a sequence, e.g. 2 + 5 + 8 + ...

Where the formula comes from

Write the sum forwards, then backwards, and add the two lines. Each pair of terms adds up to the same total, a + l. There are n such pairs, so 2Sₙ = n(a + l) and Sₙ = (n/2)(a + l). Putting l = a + (n − 1)d gives Sₙ = (n/2)[2a + (n − 1)d]. This is the idea behind the famous quick sum 1 + 2 + ... + 100 = 5050.

Example 1: sum of an AP

Problem

Find the sum of the first 30 terms of the AP 5, 9, 13, ....

Working

a = 5, d = 4, n = 30.
Formula: Sₙ = (n/2)[2a + (n − 1)d].
S₃₀ = (30/2)[2 × 5 + (30 − 1) × 4]
= 15 × (10 + 116) = 15 × 126 = 1890.
Check: t₃₀ = 5 + 29 × 4 = 121, and (30/2)(5 + 121) = 15 × 126 = 1890.

Answer

S₃₀ = 1890

Example 2: how many terms give a sum

Problem

How many terms of the series 2 + 6 + 10 + ... make 200?

Working

a = 2, d = 4. Put Sₙ = 200.
Formula: Sₙ = (n/2)[2a + (n − 1)d].
(n/2)[2 × 2 + (n − 1) × 4] = 200
(n/2)(4n) = 200, so 2n² = 200.
n² = 100, so n = 10 (n cannot be negative).
Check: t₁₀ = 38, and (10/2)(2 + 38) = 200.

Answer

10 terms

Example 3: sum of multiples in a range

Problem

Find the sum of all natural numbers between 100 and 200 that are divisible by 9.

Working

The numbers are 108, 117, ..., 198: an AP with a = 108, d = 9, l = 198.
Formulas: n = (l − a)/d + 1 and Sₙ = (n/2)(a + l).
n = (198 − 108)/9 + 1 = 10 + 1 = 11.
S₁₁ = (11/2)(108 + 198) = (11/2) × 306
= 11 × 153 = 1683.

Answer

The sum is 1683.

Example 4: a word problem

Problem

Meena saves ₹ 50 in the first week and ₹ 10 more each week than the week before. How much has she saved in 20 weeks?

Working

Weekly savings: 50, 60, 70, ..., an AP with a = 50, d = 10, n = 20.
Formula: Sₙ = (n/2)[2a + (n − 1)d].
S₂₀ = (20/2)[2 × 50 + 19 × 10]
= 10 × (100 + 190) = 10 × 290 = 2900.

Answer

She has saved ₹ 2900.

In real life

A tiffin centre makes 5 more idlis each day than the day before. Sₙ gives its total for the month without adding 30 numbers by hand.

Exam angle

2 marks: “Find the sum of 4 + 9 + 14 + ... to 15 terms.” a = 4, d = 5. S₁₅ = (15/2)[8 + 14 × 5] = (15/2) × 78 = 585.

2 marks: “If Sₙ = 3n² + 2n for an AP, find the first term and the common difference.” a = S₁ = 5. S₂ = 16, so t₂ = 16 − 5 = 11 and d = 11 − 5 = 6.

5 marks: “A school hall has 25 seats in the first row, 28 in the second, 31 in the third, and so on, for 20 rows. How many seats are there in all?” AP with a = 25, d = 3, n = 20. S₂₀ = 10[50 + 19 × 3] = 10 × 107 = 1070 seats. Check with l = 82: 10 × (25 + 82) = 1070.

Remember it

Pair the ends: first + last, times half the number of terms.