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.
Students who can do this for one question but hesitate on an unfamiliar wording often gain from watching someone else think aloud. Our teachers do this in online one-to-one Additional Mathematics tuition.