Lesson 09 · Geometric Sequences

Worksheet · Mia
Arithmetic sequences add the same amount each step. Geometric sequences multiply by the same amount — called the common ratio. Doubling bacteria, halving medication in the bloodstream, tripling followers — these are all geometric. The key test: divide any term by the previous one. If you always get the same number, the ratio is constant and the sequence is geometric.
Arithmetic — add $d$
3 +5 8 +5 13 +5 18
Formula: $a_n = a_1 + (n-1)d$
Geometric — multiply by $r$
3 ×3 9 ×3 27 ×3 81
Formula: $a_n = a_1 \cdot r^{n-1}$
Formula for the $n$th term of a geometric sequence: $$a_n = a_1 \cdot r^{n-1}$$ $a_1$ = first term, $r$ = common ratio, $n$ = position.
Finding $r$: divide any term by the one before it. $\dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = r$.
If $r > 1$: growing sequence.   If $0 < r < 1$: shrinking sequence.   If $r < 0$: alternating signs.

Worked Examples

Worked Example 1 — identify and extend

A social media post about Copper goes viral. On day 1 it gets 6 views. Each day the views triple.

6 ×3 18 ×3 54 ×3 162 ×3 486

$a_1 = 6$, $r = 3$. Views on day 7:

$$a_7 = 6 \cdot 3^{7-1} = 6 \cdot 3^6 = 6 \cdot 729 = 4\,374 \text{ views}$$
Worked Example 2 — shrinking ratio

A vet gives Copper a dose of 96 mg of a supplement. The body absorbs half of whatever remains each day. List the first 5 days and find the formula.

96 ×½ 48 ×½ 24 ×½ 12 ×½ 6
$$a_n = 96 \cdot \left(\tfrac{1}{2}\right)^{n-1}$$

The amount halves every day but never quite reaches zero.

Worked Example 3 — find the ratio from two terms

A geometric sequence has $a_1 = 5$ and $a_4 = 135$. Find $r$.

$$a_4 = a_1 \cdot r^{4-1} \quad\Rightarrow\quad 135 = 5 \cdot r^3 \quad\Rightarrow\quad r^3 = 27 \quad\Rightarrow\quad r = 3$$

Practice Problems

Problems 1–3: find $r$ and fill in the missing term(s).

  1. Fan club. The number of members in Copper's online fan club each month: 5, 15, 45, __, 405, __
    $r = $ Missing terms:
  2. Antibiotic. A horse antibiotic dose in mg remaining in the bloodstream each hour: 256, 128, 64, __, 16, __
    $r = $ Missing terms:
  3. Decide: arithmetic or geometric? Classify each sequence. If geometric, state $r$; if arithmetic, state $d$.
    (a) 7, 14, 21, 28, …
    (b) 1, 2, 4, 8, 16, …
    (c) 100, 10, 1, 0.1, …
    (d) 50, 43, 36, 29, …

Problems 4–6: use the $n$th term formula.

  1. Sponsorship deal. Mia gets a sponsorship where her monthly payment doubles each month, starting at 25 €.
    (a) Write the formula $a_n$.
    (b) How much does she receive in month 7?
    (c) In which month does the payment first exceed 1 000 €? (Try values of $n$.)
  2. Fence post rust. A fence post loses $\frac{1}{3}$ of its remaining strength each year from rust. It starts at 810 units of strength.
    Write the formula:
    Strength in year 5:
    In which year does the strength first fall below 10 units?
  3. Find the ratio. A geometric sequence has $a_1 = 2$ and $a_5 = 162$. Find $r$, then write $a_n$.
From here — write the sequence and formula yourself.
  1. Training intensity. Each week Mia increases the distance she rides by 20%. She starts at 15 km in week 1.
    Write the formula ($r = 1.2$):
    Distance in week 6 (round to 1 decimal place):
  2. Rumour at the yard. On Monday Mia tells 1 friend a piece of news. Each day, every person who knows tells 2 new people. How many new people hear the news on day 6?
    Write the sequence first, then use the formula.
  3. Mixed review. Mia has two investment options for her prize money: Write the $n$th term formula for each:
    Find the value after 10 years for each (round Option B to the nearest €):
    Which is better after 10 years? After 20 years?
  4. Challenge — find $a_1$ and $r$. The 2nd term of a geometric sequence is 12 and the 5th term is 96. Find $r$, then find $a_1$.
    Hint: use $\dfrac{a_5}{a_2} = r^3$.
What's next → You've now seen both sequence types. Lesson 10 brings them together with a mixed review and introduces slope — the rate of change of a line on a graph. You'll see that an arithmetic sequence is really just a line in disguise: the common difference is the slope.
Show answers
  1. $r = 3$; missing terms: 135 and 1215
  2. $r = \tfrac{1}{2}$; missing terms: 32 and 8
  3. (a) Arithmetic, $d = 7$; (b) Geometric, $r = 2$; (c) Geometric, $r = 0.1$; (d) Arithmetic, $d = -7$
  4. (a) $a_n = 25 \cdot 2^{n-1}$; (b) $a_7 = 25 \cdot 64 = 1\,600$ €; (c) $a_6 = 25 \cdot 32 = 800$ € (no), $a_7 = 1\,600$ € (yes) — month 7
  5. $a_n = 810 \cdot \left(\tfrac{2}{3}\right)^{n-1}$; $a_5 = 810 \cdot \left(\tfrac{2}{3}\right)^4 = 810 \cdot \tfrac{16}{81} = 160$ units; $a_6 = \approx 106.7$, $a_7 \approx 71.1$, $a_8 \approx 47.4$, $a_9 \approx 31.6$, $a_{10} \approx 21.1$, $a_{11} \approx 14.1$, $a_{12} \approx 9.4$ — year 12
  6. $162 = 2 \cdot r^4 \Rightarrow r^4 = 81 \Rightarrow r = 3$; $a_n = 2 \cdot 3^{n-1}$
  7. $a_n = 15 \cdot 1.2^{n-1}$; $a_6 = 15 \cdot 1.2^5 = 15 \cdot 2.48832 \approx 37.3$ km
  8. Sequence of new people: 1, 2, 4, 8, 16, 32; $a_n = 1 \cdot 2^{n-1}$; day 6: $2^5 = 32$ new people
  9. A: $a_n = 200 + (n-1) \cdot 30 = 170 + 30n$; B: $a_n = 200 \cdot 1.1^{n-1}$; after 10 years: A = $170 + 300 = 470$ €, B = $200 \cdot 1.1^9 \approx 200 \cdot 2.358 \approx 472$ € — roughly equal; after 20 years: A = $170 + 600 = 770$ €, B = $200 \cdot 1.1^{19} \approx 200 \cdot 6.116 \approx 1\,223$ € — B is significantly better
  10. $\dfrac{a_5}{a_2} = r^3 = \dfrac{96}{12} = 8 \Rightarrow r = 2$; $a_2 = a_1 \cdot 2 = 12 \Rightarrow a_1 = 6$
← All lessons
Focus 25:00

Ask your tutor