Year 11 Extended · Unit 3 Sequences, series, formulae and proportion · Lesson 1 of 7

Why can a long list of numbers be added in one step?

A sequence that adds the same amount each time can be summed without adding term by term. This lesson builds the formula from a picture and then uses it backwards.

Number and AlgebraNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: find the 10th term of 4, 7, 10, 13, …
    Answer31. 4 + 9 × 3.
  2. From Year 9: solve 3x + 5 = 29.
    Answerx = 8.
  3. From Year 7: work out (7 + 43) × 6 ÷ 2.
    Answer150.
Words used in this lesson
series
the sum of the terms of a sequence. For example: 4 + 8 + 12.
partial sum
the sum of only the first n terms. For example: S3.
first term
the term in position 1, written a. For example: a = 6.
common difference
the amount added to get from each term to the next, written d. For example: d = −2.
last term
the term in position n, equal to a + (n − 1)d. For example: the 5th of 5.

Can the seats be counted without adding fifteen numbers?

A hall in Kumasi has 15 rows. The front row has 12 seats and every row behind has 4 more seats than the row in front. The school needs the total number of seats, and adding fifteen numbers one at a time is slow and invites slips.

Predict first: imagine the list of seat numbers written once forwards and once backwards, one above the other. What do you expect about the total of each pair in a column? Then use the figure: change the first term and the difference and watch the column heights.

3
2
6
16each column adds to
48sum of the terms

First term 3, difference 2, 6 terms: every column adds to 16, so the 6 columns hold 96 and the sum of the terms is 96 ÷ 2 = 48.

Show me the steps
  1. First term 3, difference 2, 6 terms (3, 5, 7, 9, 11, 13): every column adds to 16, so the sum is 6 × 16 ÷ 2 = 48.
  2. First term 1, difference 0, 5 terms (1, 1, 1, 1, 1): every column adds to 2, so the sum is 5 × 2 ÷ 2 = 5.
  3. First term 5, difference 4, 4 terms (5, 9, 13, 17): every column adds to 22, so the sum is 4 × 22 ÷ 2 = 44.
  4. Any first term a, difference d and n terms: every column adds to 2a + (n − 1)d, so the sum is n ÷ 2 × (2a + (n − 1)d).

Write the sum twice, once reversed. The first column adds the first and last terms, and moving one place along raises one number by d and lowers its partner by d, so every column adds to the same total, 2a + (n − 1)d. There are n columns and they hold the sum twice, so

Sn = n ÷ 2 × (2a + (n − 1)d) = n ÷ 2 × (first term + last term).

The last term of n terms is a + (n − 1)d, because the first term has no differences added to it.

Two common slips. Using a + nd as the last term: for 10, 20, 30, 40 the last term is 10 + 3 × 10, not 10 + 4 × 10. And stopping at n × (first + last): that is twice the sum, so it must be halved.

Worked examples

1. Find the sum of the first 20 terms of 5 + 9 + 13 + …

  1. The first term is 5 and the difference is 4. The 20th term is 5 + 19 × 4 = 81.
  2. S20 = 20 ÷ 2 × (5 + 81) = 10 × 86 = 860.

2. Add the multiples of 7 from 7 to 700.

  1. The first term is 7 and the last is 700. The number of terms is 700 ÷ 7 = 100.
  2. The sum is 100 ÷ 2 × (7 + 700) = 50 × 707 = 35 350.

3. How many terms of 3 + 8 + 13 + … give a sum of 255?

  1. Sn = n ÷ 2 × (6 + 5(n − 1)) = n ÷ 2 × (5n + 1). Set it equal to 255: 5n2 + n = 510.
  2. 5n2 + n − 510 = 0 factorizes as (n − 10)(5n + 51) = 0. A number of terms cannot be negative, so n = 10.

Your go. The first step is given. Find the sum of 40 + 37 + 34 + … for 12 terms. The 12th term is 40 + 11 × (−3) = 7.

  1. The first term is 40 and the last term is 7.
  2. Use Sn = n ÷ 2 × (first + last) with n = 12.
Check the answer

12 ÷ 2 × (40 + 7) = 6 × 47 = 282.

Check before you go on

1. An arithmetic sequence has 8 terms. The first term is 6 and the last term is 41. What is the sum of the 8 terms?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Find the sum of the first 10 terms of the sequence 9, 13, 17, 21, …

3. A farm in Punjab plants trees in rows. The first row has 6 trees and each row after it has 3 more than the one before. There are 14 rows. How many trees are there in total?

Level 2 · Building

4. The first term of an arithmetic sequence is 4 and the sum of its first 10 terms is 220. What is the common difference?

5. Find the sum 2 + 5 + 8 + … + 98.

Level 3 · Stretch

6. In an arithmetic sequence the 3rd term is 11 and the 8th term is 31. Find the sum of the first 20 terms.

7. The sum of the first 10 terms of an arithmetic sequence is 185 and the sum of the first 20 terms is 670. Find the first term.

Something to think about

The pairing trick sets two copies of the sequence side by side. Does it still work when the number of terms is odd, so that one term has no partner?

One way to think about it

Yes. The reversed copy is a second full list, so there are n columns whatever n is, and the middle term of an odd list is just paired with itself. Twice the sum is n columns of the same total, and halving gives the sum.

Exit ticket

8. What is the sum of the first 30 positive even numbers, 2 + 4 + 6 + … + 60?