In a geometric progression each term is the previous term multiplied by the same number r. To recover r, divide a later term by an earlier one and take the root that matches the gap between them. One case needs extra care: an even gap gives two possible signs.
This lesson builds on finding terms from two conditions and leads into sums to infinity, where the size of r decides everything.
How do you get r from two terms?
The nth term is arn−1. So the ratio of any two terms cancels a:
(term at position q) ÷ (term at position p) = rq−p
The exponent is the gap between positions. Take the root that matches that gap: a cube root if the gap is three, a square root if it is two.
What if three terms are given in progression?
If three consecutive terms are p, q and s, then q ÷ p = s ÷ q, so q² = ps. When the terms contain an unknown x, this gives an equation in x. Solve it, substitute to get the terms, then read off r.
Worked example
A geometric progression has 3rd term 12 and 6th term 96. Find r, the first term and the 8th term.
Step 1, equations: ar2 = 12 and ar5 = 96.
Step 2, divide: ar5 ÷ ar2 = 96 ÷ 12, so r3 = 8 and r = 2.
Step 3, find a: a × 22 = 12, so 4a = 12 and a = 3.
Step 4, check: 3rd term = 3 × 4 = 12. 6th term = 3 × 32 = 96. Both agree.
Step 5, answer: the 8th term is 3 × 27 = 3 × 128 = 384.
So r = 2, a = 3 and the 8th term is 384.
The mistake to watch for
A cube root has only one real answer, but a square root has two. The slip is to give only the positive root when the gap is even.
Mistaken working: 2nd term 12 and 4th term 48. Then r2 = 4, so r = 2 and a = 6.
The value r = −2 also satisfies r2 = 4 and was dropped.
The correction is to write r = ±2 and test both.
If r = 2 then a = 6, and the terms are 6, 12, 24, 48. If r = −2 then a = −6, and the terms are −6, 12, −24, 48. Both have 2nd term 12 and 4th term 48, so both are valid unless the question adds a condition such as “all terms are positive”.
Check yourself
Try these on paper, then open each answer.
1. A geometric progression has 2nd term 10 and 3rd term 4. Find r and the first term.
Show answer
r = 4 ÷ 10 = 2/5. Then a = 10 ÷ (2/5) = 10 × 5/2 = 25.
Check: 25, 10, 4 is correct, since 25 × 2/5 = 10 and 10 × 2/5 = 4. So r = 2/5 and a = 25.
2. The numbers x + 1, x + 4 and x + 10 are consecutive terms of a geometric progression. Find x and the common ratio.
Show answer
(x + 4)² = (x + 1)(x + 10), so x² + 8x + 16 = x² + 11x + 10. This gives 6 = 3x, so x = 2.
The terms are 3, 6, 12, so r = 2. Check: 6 ÷ 3 = 2 and 12 ÷ 6 = 2.
3. A geometric progression of positive terms has 3rd term 18 and 5th term 8. Find r and the first term.
Show answer
r² = 8 ÷ 18 = 4/9, so r = ±2/3. Positive terms need a positive ratio, so r = 2/3.
Then a × 4/9 = 18 gives a = 18 × 9/4 = 81/2 = 40.5. Check: 40.5, 27, 18, 12, 8 and each step multiplies by 2/3.
Where this leads next
With r in hand, you can decide whether a geometric series has a finite total in using an infinite sum only when its condition holds. The sequence and series laboratory lets you change r and see how the terms behave, including when the sign alternates.
If you tend to drop the negative root or forget to test both signs, a teacher in online one-to-one Additional Mathematics tuition can check your root-taking on your own questions.