Chapter 6.6 — Exercise 6.5 — nth Term of GP

The Formula

Building up the pattern term by term makes the formula obvious rather than something to memorise blindly: a₁ = a, a₂ = ar, a₃ = a₂·r = ar², a₄ = a₃·r = ar³, continuing on in this same way — each new term multiplies the previous one by r exactly once, so the exponent on r simply counts how many multiplications have happened since the start.

From the general form a, ar, ar², ar³, …, the exponent on r always matches one less than the term's position — exactly the same pattern that produced the AP formula, with multiplication in place of addition:

The structural parallel between the two formulas is worth stating explicitly, since it makes both easier to remember together: aₙ = a + (n−1)d builds a term by adding d a total of (n−1) times, while aₙ = a·r^(n−1) builds a term by multiplying by r that same number of times. Both formulas encode the identical idea — "how many steps have happened since the first term" — just translated into the two different operations, addition and multiplication, that define each kind of progression.

Four GPs, Formula Found

The second row is worth a direct check: with r=−3, the sign of each term alternates automatically, since (−3) raised to an even power is positive and to an odd power is negative — the formula handles the alternation without needing any special-case treatment. The third row is worth a second look too: a=−1 combined with r=3 produces aₙ=−3ⁿ⁻¹, which looks almost identical to the formula for a plain power of 3, but the leading negative sign belongs to the coefficient a, not to the base being raised to a power — a distinction that matters the moment you need to evaluate the formula at a specific n rather than just state it.

The nth Term, in General

For the GP 5, 25, 125, … (a=5, r=5), the 10th term is a₁₀ = 5·5⁹ = 5¹⁰, and more generally, aₙ = 5·5ⁿ⁻¹ = 5ⁿ — the formula collapses neatly here because the first term and common ratio happen to be the same number, so the leading 5 out front combines with the exponent's own 5ⁿ⁻¹ to simplify the whole thing by exactly one power. This kind of simplification isn't guaranteed for every GP, only this specific one; it only happens when a and r coincide, which is why it's genuinely worth spotting when it does rather than expecting it as a general rule.

Finding a Specific Term

• a₁=9, r=1/3, find a₇: a₇ = 9·(1/3)⁶ = 9·(1/729) = 9/729 = 1/81.
• a₁=−12, r=1/3, find a₆: a₆ = −12·(1/3)⁵ = −12·(1/243) = −12/243 = −4/81.

Which Term Equals This Value?

Running the formula in reverse — exactly as Exercise 6.2 did for APs — finds a term's position from its value instead of working forward from a known position to an unknown value.

• GP 2, 8, 32, …, r=4: solving 2·4ⁿ⁻¹=512 gives 4ⁿ⁻¹=512/2=256=4⁴, so n−1=4 — the 5th term is 512.
• GP √3, 3, 3√3, …, r=√3: solving (√3)ⁿ=729=3⁶ means (√3)ⁿ=(√3)¹² after converting the base, so n=12 — the 12th term is 729. Converting 729 to a power of √3 rather than 3 directly is the key step here, since the GP's ratio is itself irrational.
• GP 1/3, 1/9, 1/27, …, r=1/3: solving (1/3)ⁿ=1/2187=(1/3)⁷ gives n=7 directly, without needing any conversion between bases — the 7th term is 1/2187.

Jumping Between Terms Without the Start

Given that the 8th term of a GP is 192 and r=2, finding the 12th term doesn't require recovering a first: the ratio between any two terms four positions apart is always r⁴, so a₁₂ = a₈·r⁴ = 192×16 = 3072. This shortcut — multiplying by rᵏ to jump k positions from any known term — works from any starting point in a GP, not just from the first term, since the ratio between aₘ and aₘ₊ₖ only ever depends on how many steps k separate them, never on where in the sequence those two terms actually sit.

It's worth seeing why this shortcut is valid rather than just accepting it. By the nth-term formula, a₁₂ = a·r¹¹ and a₈ = a·r⁷. Dividing one by the other: a₁₂/a₈ = r¹¹⁻⁷ = r⁴, and the first term a cancels out completely, since it appears in both the numerator and the denominator. That cancellation is exactly why the calculation never needed to recover a in the first place.

Reconstructing a GP From Two Terms

If the 4th term of a GP is 2/3 and the 7th term is 16/81, the ratio between them isolates r directly: a₇/a₄ = r³ = (16/81)/(2/3) = 8/27, so r = 2/3. Substituting back into a₄ = a·r³ gives a = 9/4.

Notice the exponent used to isolate r here — the gap between the 4th and 7th term is 3 positions, which is exactly why dividing produces r³ rather than any other power. The same "divide to isolate rᵏ, where k is the gap between the two known positions" technique underlies both this problem and the jump-ahead shortcut in the previous section, just applied to a pair of terms that don't happen to be the first term of the sequence.

Two GPs, Equal at the Same Term

Two separate GPs — 162, 54, 18, … (a=162, r=1/3) and 2/81, 2/27, 2/9, … (a=2/81, r=3) — happen to share a value at some term n. Setting their nth terms equal, 162·(1/3)ⁿ⁻¹ = (2/81)·3ⁿ⁻¹, and simplifying both sides into powers of the same base reduces the whole equation to 3^(2(n−1)) = 3⁸, giving 2(n−1)=8, so n = 5. The two GPs move in opposite directions entirely — one shrinking, one growing — which makes it genuinely striking that they cross paths at exactly one shared term rather than never meeting at all.

The Chapter, Complete

This exercise closes out Progressions by giving geometric progressions the same complete toolkit that arithmetic progressions received earlier in this same chapter: a recognition test (Exercise 6.4) and a direct nth-term formula, mirroring Exercise 6.1 and Exercise 6.2 almost exactly, just once addition-based and once multiplication-based throughout. For the AP sum formula this chapter covers separately (with no GP equivalent introduced here), see Exercise 6.3, or revisit both progression types from the very start in the chapter introduction. The next chapter, Coordinate Geometry, moves on from sequences of numbers entirely, turning instead to points and distances measured on a plane.