Many sequence questions give two conditions, such as “the 5th term is 17 and the 12th term is 38”, and then ask for another term or a sum. The plan is always the same: write each condition as an equation in the first term and the common difference (or ratio), solve the pair, then answer the question that was asked.
This is the opening skill of arithmetic and geometric series and it feeds every other lesson in the module.
How do two terms pin down a sequence?
An arithmetic progression is fixed by its first term a and common difference d. The nth term is a + (n − 1)d. A geometric progression is fixed by a and the common ratio r, and its nth term is arn−1.
The “minus one” matters. The first term has no d added, the second has one d, and the nth has n − 1 of them. Most errors in this topic start with that one number.
How do you turn the conditions into equations?
- Write the nth term rule for the type of progression.
- Substitute each position into the rule. The 5th term uses n = 5, so it becomes a + 4d.
- Solve the pair. For an arithmetic progression, subtract one equation from the other so that a disappears.
- Find the other unknown by substituting back.
- Check both original conditions, then answer what the question asked.
Worked example
An arithmetic progression has 5th term 17 and 12th term 38. Find the first term, the common difference and the 20th term.
Step 1, equations: a + 4d = 17 and a + 11d = 38.
Step 2, subtract: (a + 11d) − (a + 4d) = 38 − 17, so 7d = 21 and d = 3.
Step 3, find a: a + 4(3) = 17, so a = 5.
Step 4, check: 5th term = 5 + 12 = 17. 12th term = 5 + 33 = 38. Both agree.
Step 5, answer: the 20th term is 5 + 19(3) = 5 + 57 = 62.
So a = 5, d = 3 and the 20th term is 62.
The mistake to watch for
A common slip is to use n instead of n − 1, so that the 5th term is written as a + 5d.
Mistaken working: a + 5d = 17 and a + 12d = 38. Subtracting gives 7d = 21, so d = 3. Then a + 15 = 17, so a = 2.
The difference d is still correct, because both equations shifted together. The first term is wrong, and so the 20th term comes out as 2 + 19(3) = 59 instead of 62.
The fix is to ask “how many steps of d separate this term from the first term?” The 5th term is four steps from the first, never five. Checking the answer in the original sentence catches this: 2 + 4(3) = 14, which is not 17.
Check yourself
Try these on paper, then open each answer.
1. An arithmetic progression has 3rd term 11 and 8th term 31. Find the 15th term.
Show answer
a + 2d = 11 and a + 7d = 31. Subtracting gives 5d = 20, so d = 4. Then a = 11 − 8 = 3.
15th term = 3 + 14(4) = 59. Check: 3rd term = 3 + 8 = 11 and 8th term = 3 + 28 = 31.
2. A geometric progression has 2nd term 6 and 5th term 162. Find the 7th term.
Show answer
ar = 6 and ar4 = 162. Dividing, r3 = 27, so r = 3. Then a = 6 ÷ 3 = 2.
7th term = 2 × 36 = 2 × 729 = 1458. Check: ar4 = 2 × 81 = 162.
3. An arithmetic progression has 4th term 1 and 10th term −17. Which is the first term that is negative, and what is its value?
Show answer
a + 3d = 1 and a + 9d = −17. Subtracting gives 6d = −18, so d = −3, and a = 1 + 9 = 10.
The terms are 10, 7, 4, 1, −2, … so the first negative term is the 5th term, which is −2. Check: 10th term = 10 − 27 = −17.
Where this leads next
Once you can recover a and d reliably, use them to find totals in calculating a finite arithmetic sum. The sequence and series laboratory lets you test your own values, and the non-calculator working trainer builds speed on the arithmetic.
Some students can do the algebra but lose marks translating the sentence into equations. That is the kind of pattern our teachers look for in online one-to-one Additional Mathematics tuition.