Arithmetic sequences add each time. Geometric ones multiply.
Find the 10th term of
3, 6, 12, 24, …
The nth term is a·r⁽ⁿ⁻¹⁾ = 3 × 2⁹ = 3 × 512 = 1536. B uses 2¹⁰ — the exponent is always one less than the term number, because the first term has had no jumps yet.
Test the gaps first, then test the ratios.
3, 6, 12
not constant — not arithmetic
2, 2, 2
constant — geometric
a₁
3
×2 →
a₂
6
×2 →
a₃
12
×2 →
a₄
24
Divide any term by the one before it. If you always get the same number, that number is the common ratio r.
6 ÷ 3 = 2
second over first
12 ÷ 6 = 2
and again
24 ÷ 12 = 2
confirmed
first term a₁
3
common ratio r
2
term wanted n
10
Always check the ratio on more than one pair.
aₙ = a₁ · rn−1
| Term | Built as | Multiplications |
|---|---|---|
| a₁ = 3 | 3 | 0 |
| a₂ = 6 | 3 · 2 | 1 |
| a₃ = 12 | 3 · 2² | 2 |
| a₁₀ | 3 · 2⁹ | 9 |
The first term needed no multiplying at all, so the 10th needs nine — not ten.
Using r10 gives 3072, exactly double the right answer. The same n − 1 appears in the arithmetic formula, for the same reason.
a₁₀ = 3 · 210−1
the formula
= 3 · 29
the exponent first
= 3 · 512
2⁹ = 512
= 1536
the 10th term
3 · 2⁹ is not (3 · 2)⁹. Do the exponent first, then multiply by the first term.
| r | Sequence | Behaviour |
|---|---|---|
| r = 2 | 3, 6, 12, 24 | grows fast |
| r = ½ | 32, 16, 8, 4 | shrinks toward 0 |
| r = −2 | 3, −6, 12, −24 | alternates sign |
| r = 1 | 5, 5, 5, 5 | never changes |
A negative ratio flips the sign every step — a useful clue when a sequence zig-zags.
When |r| < 1 the terms shrink forever, which is what makes an infinite geometric series add up to a finite number.
Divide the later term by the earlier one. The power of r is the gap in positions.
a₅ ÷ a₂ = r³
term 2 to term 5 is three multiplications
r³ = 96 ÷ 12 = 8
divide
r = 2
cube root
a₁ = a₂ ÷ r = 12 ÷ 2 = 6
one step back
Check: 6, 12, 24, 48, 96 — the 2nd term is 12 and the 5th is 96 ✓.
If the two terms were 2 places apart, r² = 9 would give r = 3 or r = −3. Keep both unless the question says the terms are positive. An odd gap gives one ratio and keeps its sign: r³ = −64 means r = −4 only.
Inserting means is the same problem. Insert two geometric means between 5 and 40.
5, __, __, 40
two means make four terms, so 40 is a₄
r³ = 40 ÷ 5 = 8, r = 2
three multiplications from a₁ to a₄
means 10 and 20
5 × 2, then 10 × 2
Fraction r. Find the 7th term of 96, 48, 24, …
r = 48 ÷ 96 = 12
the terms halve
a₇ = 96 · (12)6 = 96 ÷ 64
six halvings
a₇ = 32
not a whole number, and that is fine
What is different: multiplying by 12 is dividing by 2, so the terms shrink. The answer can be a fraction.
Negative r. Find the 6th term of 2, −6, 18, …
r = −6 ÷ 2 = −3
the sign flips each step
a₆ = 2 · (−3)5
brackets round −3
a₆ = 2 · (−243) = −486
odd power keeps the minus
What is different: the sign of the answer comes from the power. Odd terms (a₁, a₃, a₅) share a₁’s sign; even terms have the opposite sign.
Working with 3 and adding the sign “by feel” drops the minus. Keep the brackets: (−3)5 is negative, (−3)4 is positive.
An even gap. In a geometric sequence a₃ = 20 and a₅ = 80. Find a₄.
r2 = 80 ÷ 20 = 4
two places apart
r = 2 or r = −2
both square to 4
a₄ = 20 · r = 40 or −40
20, 40, 80 or 20, −40, 80
What is different: an even power has two roots. Both sequences fit, so give both unless the question says the terms are positive.
Three terms in a row. x, x + 4 and x + 16 are consecutive terms of a geometric sequence. Find x.
(x + 4) ÷ x = (x + 16) ÷ (x + 4)
the two ratios are equal
(x + 4)² = x(x + 16)
middle² = first × last
x² + 8x + 16 = x² + 16x
the x² cancels
16 = 8x, x = 2
terms 2, 6, 18: ratio 3 both times ✓
What is different: in an arithmetic sequence the middle is the average of its neighbours; in a geometric one the middle squared is their product.
5 · 2n − 1 = 640
the formula, n unknown
2n − 1 = 128
divide by the first term
2n − 1 = 27
write 128 as a power of 2
n − 1 = 7, so n = 8
match the exponents
7 is the number of doublings. The first term needs none, so 640 is the 8th term — the same n − 1 as in the formula.
Exam items rarely hand you a₁. They give two terms (divide them; the power is the gap in positions), ask which term a number is (match the powers), or wrap the sequence in a growth story such as bacteria doubling. Ten-second check: rebuild the terms from your a₁ and r and see the given ones land in the right places.
Find the 8th term of
5, 15, 45, 135, …
5 × 3⁷ = 5 × 2,187 = 10,935. C uses 3⁸. Getting from the 1st term to the 8th takes seven multiplications, not eight.
Hint: divide to find r, then remember the exponent is one less than the term number. You want r7, not r8.
