Linear Sequences and Common Difference

Worked example

1. 1

   $6,13,20,27$ — The question: find the next term. Start by looking at how each term changes to the next.

2. 2

   $13-6=7$ — Take each term from the one after it. The gap is 7.

3. 3

   $20-13=7,27-20=7$ — Check the other gaps. They are all 7, so the rule is “add 7”.

4. 4

   $27+7=34$ — Add 7 to the last term. The next term is 34.

Worked example

1. 1

   $3,8,13,18$ — The question: find the nth term. Find the gap first, like the last card.

2. 2

   $\text{difference}=5$ — 8 − 3 = 5, and every gap is 5.

3. 3

   $5n$ — The difference goes in front of n. Why? When n goes up by 1, 5n goes up by 5, just like the sequence.

4. 4

   $5n:5,10,15,20$ — Write out 5n for n = 1, 2, 3, 4, under the sequence.

5. 5

   $3-5=-2$ — Compare the two rows: each term is 2 less than the 5n above it.

6. 6

   $\text{nth term}=5n-2$ — So take 2 away from 5n.

7. 7

   $5\times 2-2=8$ — Check with n = 2. You get 8, the 2nd term, so the answer is right.

Worked example

1. 1

   $9,13,17,21$ — The question: find the nth term.

2. 2

   $a=9,d=4$ — The first term is 9, and 13 − 9 = 4.

3. 3

   $9+(n-1)\times 4$ — Put a and d into the formula.

4. 4

   $9+4n-4$ — Expand the bracket: 4 × n = 4n and 4 × (−1) = −4.

5. 5

   $4n+5$ — Collect the numbers: 9 − 4 = 5.

6. 6

   $4\times 1+5=9$ — Check with n = 1. You get 9, the first term.

Worked example

1. 1

   $50,47,44,41$ — The question: find the nth term.

2. 2

   $a=50,d=-3$ — It goes down by 3, so d = −3.

3. 3

   $50+(n-1)\times (-3)$ — Put a and d in. The bracket keeps the minus safe.

4. 4

   $50-3n+3$ — Expand: (−3) × n = −3n and (−3) × (−1) = +3, because minus times minus is plus.

5. 5

   $53-3n$ — Collect the numbers: 50 + 3 = 53. Check n = 1: 53 − 3 = 50.

Worked example

1. 1

   $5\times 1-2=3$ — The 1st term: put n = 1. Multiply first, then take 2.

2. 2

   $5\times 2-2=8,5\times 3-2=13$ — Do the same with n = 2 and n = 3. The first three terms are 3, 8, 13.

3. 3

   $5\times 20-2=98$ — The 20th term: put n = 20. 5 × 20 = 100, then 100 − 2 = 98.

Worked example

1. 1

   $5n-2=123$ — Set the nth term equal to the number. If n comes out whole, 123 is a term.

2. 2

   $5n=125$ — Add 2 to both sides.

3. 3

   $n=25$ — Divide both sides by 5. 25 is a whole number, so yes: 123 is the 25th term.

4. 4

   $5n-2=150$ — Now test 150 the same way.

5. 5

   $5n=152$ — Add 2 to both sides.

6. 6

   $n=30.4$ — Divide by 5. There is no 30.4th term, so 150 is not in the sequence.

Worked example

1. 1

   $5n-2>100$ — Write it as an inequality: the term must be more than 100.

2. 2

   $5n>102$ — Add 2 to both sides.

3. 3

   $n>20.4$ — Divide both sides by 5.

4. 4

   $n=21$ — n must be a whole number, so take the next one up: 21. Rounding down to 20 would be too small.

5. 5

   $5\times 21-2=103$ — Put n = 21 back in. Check: n = 20 gives 98, which is not more than 100. So the answer is 103.

Worked example

1. 1

   $\text{3rd term}=11,\text{7th term}=31$ — The question: find the difference and the first term.

2. 2

   $7-3=4\text{steps}$ — From the 3rd term to the 7th term you add the difference 4 times.

3. 3

   $31-11=20$ — Over those 4 steps the terms grew by 20.

4. 4

   $d=20÷4=5$ — Every step is the same size, so share the 20 equally. The difference is 5.

5. 5

   $\text{2nd term}=11-5=6$ — Go back one step from the 3rd term: take 5 away.

6. 6

   $\text{1st term}=6-5=1$ — Back one more step. The first term is 1.

7. 7

   $1,6,11,16,21,26,31$ — Check: the 3rd term is 11 and the 7th is 31. Now the nth term is 5n − 4.

Worked example

1. 1

   $d=12-5=7$ — Find the difference from the terms you are given.

2. 2

   $27-25=2\text{steps}$ — The 27th term is 2 steps after the 25th.

3. 3

   $173+2\times 7=187$ — Add the difference once for each step. The 27th term is 187.

4. 4

   $173-2\times 7=159$ — Going back works too: the 23rd term is 2 steps before, so take 7 off twice.