What makes a sequence geometric?
Where arithmetic sequences add a fixed amount each step, geometric sequences multiply by a fixed amount each step. That fixed multiplier is called the common ratio, written $r$.
Spot the pattern
$1, \ 2, \ 4, \ 8, \ 16, \ \dots$ each term is the previous one $\times \ 2$.
$2, \ 6, \ 18, \ 54, \ \dots$ each term is the previous one $\times \ 3$:
$\dfrac{6}{2} = \dfrac{18}{6} = \dfrac{54}{18} = 3 \ = \ $ the common ratio.
The test for geometric:
$\dfrac{T_{n+1}}{T_n} = \text{constant} = r$
If the ratio comes out to a number with no $n$ in it, the sequence is geometric.
Worked example — $T_n = 5^{n}$
Prove that $T_n = 5^{n}$ defines a geometric sequence.
$T_n = 5^{n}$
$T_{n+1} = 5^{n+1}$
$\dfrac{T_{n+1}}{T_n} = \dfrac{5^{n+1}}{5^{n}} = 5^{(n+1) - n} = 5$
$5$ is a constant — no $n$. The sequence is geometric, with $r = 5$.
Two index laws are doing the heavy lifting:
$\dfrac{a^{p}}{a^{q}} = a^{p-q}$ and $a^{p+q} = a^{p} \cdot a^{q}$
You can also write $5^{n+1} = 5^{n}(5)$, which then cancels cleanly with $5^{n}$ in the denominator.
Worked example — $T_n = 2(6^{n})$
$T_{n+1} = 2(6^{n+1})$
$\dfrac{T_{n+1}}{T_n} = \dfrac{2(6^{n+1})}{2(6^{n})}$
The $2$'s cancel. So do most of the powers of 6:
$= 6^{(n+1) - n} = 6$
Constant. Geometric, $r = 6$.
A useful intuition: any $T_n$ of the form $T_n = a \cdot r^{n}$ (or $a \cdot r^{n-1}$) is automatically geometric — the constant in front cancels, and the powers subtract to leave $r$. Mixed expressions like $T_n = 3^{n} + 2^{n}$ are not geometric because the ratio doesn't simplify to a constant.
Try this
Prove that $T_n = 7^{n}$ defines a geometric sequence and state $r$.
$T_{n+1} = 7^{n+1}$
$\dfrac{T_{n+1}}{T_n} = \dfrac{7^{n+1}}{7^{n}} = 7^{(n+1) - n} = 7$
Constant ✓
Try this
Prove that $T_n = 4(3^{n})$ defines a geometric sequence and state $r$.
$T_{n+1} = 4(3^{n+1})$
$\dfrac{T_{n+1}}{T_n} = \dfrac{4(3^{n+1})}{4(3^{n})} = 3^{(n+1) - n} = 3$
Try this
Is the sequence $T_n = 2^{n} + 3$ geometric?
$T_1 = 2 + 3 = 5$
$T_2 = 4 + 3 = 7$
$T_3 = 8 + 3 = 11$
$\dfrac{T_2}{T_1} = \dfrac{7}{5} = 1.4$
$\dfrac{T_3}{T_2} = \dfrac{11}{7} \approx 1.57$
The ratios disagree, so the sequence isn't geometric.
Notation: first term $a$ and common ratio $r$
$a$ = first term: $a = T_1 = S_1$
$r$ = common ratio: $r = \dfrac{T_2}{T_1} \ = \ \dfrac{T_{n+1}}{T_n}$
Quick read on how $r$ shapes the sequence:
$r > 1$ terms grow without bound
$0 < r < 1$ terms shrink towards zero
$r < 0$ terms alternate sign as they go
Reading $a$ and $r$ off a list
(i) $2, \ 6, \ 18, \ \dots$
$a = 2, \ \ r = \dfrac{6}{2} = 3$
(ii) $\dfrac{1}{2}, \ \dfrac{3}{4}, \ \dfrac{9}{8}, \ \dots$
$a = \dfrac{1}{2}$
$r = \dfrac{T_2}{T_1} = \dfrac{3}{4} \div \dfrac{1}{2} = \dfrac{3}{4} \times \dfrac{2}{1} = \dfrac{3}{2}$
Check the next jump: $\dfrac{T_3}{T_2} = \dfrac{9}{8} \div \dfrac{3}{4} = \dfrac{9}{8} \times \dfrac{4}{3} = \dfrac{3}{2}$ ✓
The compound-interest connection
The sequence $300(1.07), \ 300(1.07)^{2}, \ 300(1.07)^{3}, \ \dots$ is geometric:
$a = 300(1.07)$
$r = 1.07$
This is exactly what compound interest does — each year multiplies last year's balance by the same factor $(1 + i)$. The compound-interest formula (in Tables) is:
$F = P(1 + i)^{t}$
$P$ = present value (principal). $F$ = future value. $i$ = interest rate as a decimal. $t$ = time.
Try this
Find $a$ and $r$ for $5, \ 15, \ 45, \ 135, \ \dots$
$a = 5$
$r = \dfrac{15}{5} = 3$
Try this
Find $a$ and $r$ for $\dfrac{2}{3}, \ \dfrac{4}{9}, \ \dfrac{8}{27}, \ \dots$
$a = \dfrac{2}{3}$
$r = \dfrac{4}{9} \div \dfrac{2}{3} = \dfrac{4}{9} \times \dfrac{3}{2} = \dfrac{12}{18} = \dfrac{2}{3}$
Try this
Find $a$ and $r$ for $8, \ -4, \ 2, \ -1, \ \dots$
$a = 8$
$r = \dfrac{-4}{8} = -\dfrac{1}{2}$
Check: $-4 \times -\dfrac{1}{2} = 2$ ✓ $2 \times -\dfrac{1}{2} = -1$ ✓
Try this
€1000 is invested at $5\%$ compound interest per year. Write the values after years 1, 2, 3 as a geometric sequence and state $a$ and $r$.
After year 1: $1000(1.05)$
After year 2: $1000(1.05)^{2}$
After year 3: $1000(1.05)^{3}$
$a = 1000(1.05), \ \ r = 1.05$
The general term: $T_n = ar^{n-1}$
Write the terms of a geometric sequence using $a$ and $r$:
$T_1 = a$
$T_2 = ar$
$T_3 = ar^{2}$
$T_4 = ar^{3}$
$\vdots$
Term $n$ has been multiplied by $r$ exactly $(n-1)$ times.
The general term (in Tables):
$T_n = ar^{n-1}$
Worked example (i)
Find $T_n$ for $5, \ 20, \ 80, \ \dots$
$a = 5, \ \ r = \dfrac{20}{5} = 4$
$T_n = ar^{n-1} = 5(4)^{n-1}$
$T_n = 5 \cdot 4^{n-1}$
Worked example (ii) — fractional $r$
Find $T_n$ for $3, \ \dfrac{3}{4}, \ \dfrac{3}{16}, \ \dots$
$a = 3, \ \ r = \dfrac{3/4}{3} = \dfrac{1}{4}$
$T_n = 3 \left(\dfrac{1}{4}\right)^{n-1}$
$= 3 \cdot \dfrac{1}{4^{n-1}}$
$T_n = \dfrac{3}{4^{n-1}}$
Which term has a given value? (Powers $\Rightarrow$ Logs)
Which term of the sequence $3, \ 15, \ 75, \ \dots$ has the value $1{,}171{,}875$?
$a = 3, \ \ r = 5$
Set $T_n = 1{,}171{,}875$:
$3(5)^{n-1} = 1{,}171{,}875$
$5^{n-1} = 390{,}625$
Spot-the-power route. If you recognise $390{,}625 = 5^{8}$:
$5^{n-1} = 5^{8}$
$n - 1 = 8$
$n = 9$
Don't recognise the power? Take logs of both sides:
$\log(5^{n-1}) = \log(390{,}625)$
$(n-1)\log 5 = \log(390{,}625)$
$n - 1 = \dfrac{\log(390{,}625)}{\log 5} = 8$
$n = 9$
Recall from logs: $b^{p} = n \ \Leftrightarrow \ \log_b n = p$. So $5^{n-1} = 390{,}625$ gives $n - 1 = \log_{5}(390{,}625) = 8$ in one step.
The 9th term equals $1{,}171{,}875$.
Try this
Find $T_n$ for $2, \ 6, \ 18, \ 54, \ \dots$ and use it to find $T_{10}$.
$a = 2, \ \ r = 3$
$T_n = 2 \cdot 3^{n-1}$
$T_{10} = 2 \cdot 3^{9} = 2 \cdot 19{,}683$
Try this
Find $T_n$ for $\dfrac{1}{2}, \ \dfrac{1}{6}, \ \dfrac{1}{18}, \ \dots$
$a = \dfrac{1}{2}$
$r = \dfrac{1/6}{1/2} = \dfrac{1}{6} \times \dfrac{2}{1} = \dfrac{1}{3}$
$T_n = \dfrac{1}{2}\left(\dfrac{1}{3}\right)^{n-1}$
Try this
Which term of $2, \ 6, \ 18, \ \dots$ has the value $4374$?
$a = 2, \ \ r = 3$
$2 \cdot 3^{n-1} = 4374$
$3^{n-1} = 2187$
Powers of 3: $3^{7} = 2187$.
$n - 1 = 7 \ \Rightarrow \ n = 8$
Try this
In a geometric sequence $a = 2$ and $r = 5$. Use logs to find $n$ such that $T_n = 1{,}250{,}000$.
$2 \cdot 5^{n-1} = 1{,}250{,}000$
$5^{n-1} = 625{,}000$
$n - 1 = \log_{5}(625{,}000)$
$= \dfrac{\log(625{,}000)}{\log 5} = 8.293\dots$
This isn't a whole number — the value $1{,}250{,}000$ is not a term of this sequence.
The series sum: $S_n = \dfrac{a(1 - r^{n})}{1 - r}$
Sum of a GP (in Tables):
$S_n = \dfrac{a(1 - r^{n})}{1 - r}$
Valid as long as $r$ is not $1$ (otherwise the denominator is zero — and the series is just $a + a + a + \dots$ anyway).
Worked example — $S_6$ and $S_{30}$
Find $S_6$ and $S_{30}$ of $2 + 4 + 8 + 16 + \dots$
$a = 2, \ \ r = 2$
$S_6 = \dfrac{2(1 - 2^{6})}{1 - 2} = \dfrac{2(1 - 64)}{-1} = \dfrac{2(-63)}{-1}$
$= 126$
$S_{30} = \dfrac{2(1 - 2^{30})}{1 - 2} = \dfrac{2(1 - 2^{30})}{-1} = 2(2^{30} - 1)$
$= 2^{31} - 2$
$S_6 = 126, \ \ S_{30} = 2^{31} - 2$
When $r > 1$ the sum is huge; leaving it as $2^{31} - 2$ is usually cleaner than the decimal.
Worked example — compound interest with monthly payments
€500 is invested at the start of each month for $12$ months. The interest rate is $0.2\%$ per month. Find the value of the investment at the end of $1$ year.
Each lodgement grows by $(1 + i) = 1.002$ for as long as it sits in the account. The first €500 sits for $12$ months, the second for $11$, and so on. The last €500 (start of month 12) sits for $1$ month.
Total value:
$500(1.002)^{12} + 500(1.002)^{11} + \dots + 500(1.002)$
$= 500\bigl[ 1.002 + (1.002)^{2} + \dots + (1.002)^{12} \bigr]$
The bracket is a GP with $a = 1.002, \ r = 1.002, \ n = 12$.
$S_{12} = \dfrac{1.002\bigl(1 - 1.002^{12}\bigr)}{1 - 1.002}$
$= 12.157\dots$
Value $= 500 \times 12.157 = $ €6078.57
Worked example — find $n$ from the last term, then sum
Find $3 + 6 + 12 + \dots + 1536$.
$1536$ is the last term, not the number of terms. Find $n$ first.
$a = 3, \ \ r = 2$
$T_n = 1536: \ 3(2)^{n-1} = 1536$
$2^{n-1} = 512 = 2^{9}$
$n - 1 = 9 \ \Rightarrow \ n = 10$
Now sum the first 10 terms:
$S_{10} = \dfrac{3(1 - 2^{10})}{1 - 2} = \dfrac{3(1 - 1024)}{-1} = 3(1023)$
$S_{10} = 3069$
Try this
Find $S_8$ of $1 + 3 + 9 + 27 + \dots$
$S_8 = \dfrac{1(1 - 3^{8})}{1 - 3} = \dfrac{1 - 6561}{-2}$
$= \dfrac{-6560}{-2} = 3280$
Try this
Find $4 + 8 + 16 + \dots + 2048$.
$a = 4, \ \ r = 2$
$4(2)^{n-1} = 2048 \ \Rightarrow \ 2^{n-1} = 512 = 2^{9}$
$n - 1 = 9 \ \Rightarrow \ n = 10$
$S_{10} = \dfrac{4(1 - 2^{10})}{1 - 2} = \dfrac{4(1 - 1024)}{-1} = 4(1023)$
Try this
€200 is invested at the start of each month for $6$ months. The interest is $0.5\%$ per month. Find the value at the end of $6$ months. Give your answer to the nearest cent.
Total $= 200\bigl[(1.005) + (1.005)^{2} + \dots + (1.005)^{6}\bigr]$
$S_6 = \dfrac{1.005(1 - 1.005^{6})}{1 - 1.005}$
$1.005^{6} \approx 1.030378$
$S_6 \approx \dfrac{1.005(-0.030378)}{-0.005} \approx 6.1076$
Total $\approx 200 \times 6.1076 \approx 1221.51$
Sim Eqs — find $a$ and $r$ from two facts
Like with arithmetic, two unknowns need two facts. The mechanics are different here because the equations are multiplicative — division and substitution rather than elimination.
Worked example 1 — two known terms
In a geometric progression $T_2 = 6$ and $T_5 = 162$. Find $a$ and $r$.
Use $T_n = ar^{n-1}$ on each:
$T_2 = ar = 6$
$T_5 = ar^{4} = 162$
Divide the second by the first — the $a$'s cancel:
$\dfrac{ar^{4}}{ar} = \dfrac{162}{6}$
$r^{3} = 27$
$r = 3$
Back-substitute: $a(3) = 6 \ \Rightarrow \ a = 2$
$a = 2, \ \ r = 3$
Another way to read it: since $T_5 = T_2 \times r^{3}$ (three steps from term 2 to term 5), you can write $r^{3} = T_5 / T_2 = 162/6 = 27$ directly. Same algebra, different angle.
Worked example 2 — a term and a series sum
In a geometric series $T_3 = 96$ and $S_2 = 30$. Find $a$ and $r$.
From $T_3$: $ar^{2} = 96 \ \Rightarrow \ a = \dfrac{96}{r^{2}}$ (1)
For $S_2$, the formula is $\dfrac{a(1 - r^{2})}{1 - r}$. But $1 - r^{2}$ factors:
$\dfrac{a(1 - r)(1 + r)}{1 - r} = a(1 + r)$
So: $a(1 + r) = 30$ (2)
The factoring trick $1 - r^{n} = (1 - r)(1 + r + r^{2} + \dots + r^{n-1})$ for $n = 2$ collapses $S_2$ to the much-friendlier $a(1 + r)$. Worth knowing.
Sub (1) into (2):
$\dfrac{96}{r^{2}}(1 + r) = 30$
$96(1 + r) = 30 r^{2}$
$16(1 + r) = 5 r^{2}$
$5r^{2} - 16r - 16 = 0$
Factor (split the middle: $-20r + 4r$):
$5r^{2} - 20r + 4r - 16 = 0$
$5r(r - 4) + 4(r - 4) = 0$
$(r - 4)(5r + 4) = 0$
$r = 4 \ \text{ or } \ r = -\dfrac{4}{5}$
If $r = 4$: $a = \dfrac{96}{16} = 6$
If $r = -\dfrac{4}{5}$: $a = \dfrac{96}{16/25} = 96 \times \dfrac{25}{16} = 150$
Two solutions: $(a, r) = (6, \ 4)$ or $(a, r) = \left(150, \ -\dfrac{4}{5}\right)$
Try this
In a GP, $T_3 = 18$ and $T_6 = 486$. Find $a$ and $r$.
$ar^{2} = 18, \ \ ar^{5} = 486$
$\dfrac{ar^{5}}{ar^{2}} = \dfrac{486}{18}$
$r^{3} = 27 \ \Rightarrow \ r = 3$
$a(9) = 18 \ \Rightarrow \ a = 2$
Try this
In a GP, $T_4 = 24$ and $T_7 = 192$. Find $a$ and $r$.
$ar^{3} = 24, \ \ ar^{6} = 192$
$r^{3} = \dfrac{192}{24} = 8 \ \Rightarrow \ r = 2$
$a(8) = 24 \ \Rightarrow \ a = 3$
Try this
In a GP, $T_2 = 12$ and $S_2 = 18$. Find $a$ and $r$.
$ar = 12$ (1)
$S_2 = a + ar = a(1 + r) = 18$ (2)
From (2): $a + ar = 18$. Sub $ar = 12$: $a + 12 = 18 \ \Rightarrow \ a = 6$
From (1): $6r = 12 \ \Rightarrow \ r = 2$
The $x$-question (geometric version)
Three consecutive terms of a geometric sequence have a constant ratio, not a constant difference. So the rule is:
$x$-question rule for GP:
$\dfrac{T_2}{T_1} = \dfrac{T_3}{T_2}$
Same ratio either side.
Worked example
Three consecutive terms of a geometric sequence are $x - 1, \ x + 3, \ x + 8$. Find $x$.
$\dfrac{x + 3}{x - 1} = \dfrac{x + 8}{x + 3}$
Cross-multiply (and don't be shy about it — both sides are just fractions):
$(x + 3)(x + 3) = (x + 8)(x - 1)$
$x^{2} + 6x + 9 = x^{2} + 7x - 8$
$9 + 8 = 7x - 6x$
$17 = x$
$x = 17$
Neat side-effect: the middle term squared equals the product of the outer two. That is, $(x+3)^{2} = (x-1)(x+8)$. This is sometimes called the geometric mean relationship, and you can use it as a one-liner.
Try this
$2, \ x, \ 18$ are three consecutive terms of a geometric sequence. Find the possible values of $x$.
$\dfrac{x}{2} = \dfrac{18}{x}$
$x^{2} = 36$
$x = 6 \ \text{ or } \ x = -6$
(With $x = 6$: $r = 3$. With $x = -6$: $r = -3$. Both valid GPs.)
Try this
$x + 2, \ x + 6, \ x + 14$ are three consecutive terms of a geometric sequence. Find $x$.
$(x + 6)^{2} = (x + 2)(x + 14)$
$x^{2} + 12x + 36 = x^{2} + 16x + 28$
$36 - 28 = 16x - 12x$
$8 = 4x$
Symmetric variables — 3 consecutive GP terms
For arithmetic we used $x - y, \ x, \ x + y$ (additive symmetry). For geometric we want multiplicative symmetry — divide and multiply by the same thing:
Symmetric labels for a GP:
$\dfrac{x}{y}, \ \ x, \ \ xy$
The ratio is $y$ either way. The product is $x^{3}$ — the $y$'s cancel automatically.
Worked example
Find three terms of a geometric sequence which add to $14$ and multiply to $64$.
Call the three terms $\dfrac{x}{y}, \ x, \ xy$.
Product $= 64$:
$\dfrac{x}{y} \cdot x \cdot xy = x^{3} = 64$
$x = 4$
Sum $= 14$:
$\dfrac{x}{y} + x + xy = 14$
Sub $x = 4$:
$\dfrac{4}{y} + 4 + 4y = 14$
Multiply through by $y$:
$4 + 4y + 4y^{2} = 14y$
$4y^{2} - 10y + 4 = 0$
$2y^{2} - 5y + 2 = 0$
$(2y - 1)(y - 2) = 0$
$y = \dfrac{1}{2} \ \text{ or } \ y = 2$
If $x = 4, y = 2$: terms are $\dfrac{4}{2}, \ 4, \ 4(2) = 2, \ 4, \ 8$.
If $x = 4, y = \dfrac{1}{2}$: terms are $\dfrac{4}{1/2}, \ 4, \ 4 \cdot \dfrac{1}{2} = 8, \ 4, \ 2$.
$2, \ 4, \ 8$ (or $8, \ 4, \ 2$ reversed).
Try this
Three consecutive terms of a geometric sequence have product $216$ and sum $26$. Find the terms.
Product: $x^{3} = 216 \ \Rightarrow \ x = 6$
Sum: $\dfrac{6}{y} + 6 + 6y = 26$
$\dfrac{6}{y} + 6y = 20$
$6 + 6y^{2} = 20y$
$6y^{2} - 20y + 6 = 0$
$3y^{2} - 10y + 3 = 0$
$(3y - 1)(y - 3) = 0 \ \Rightarrow \ y = \dfrac{1}{3} \ \text{ or } \ y = 3$
Sum to infinity — $S_{\infty}$
What happens if we just keep adding terms forever? Two scenarios:
$1 + 2 + 3 + \dots \ = \ \infty$ (terms grow — sum runs away)
$2 + 4 + 8 + 16 + \dots \ = \ \infty$ (terms grow — same fate)
$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots \ = \ 1$ (terms shrink fast enough — sum settles)
The third one is the magic case. The terms shrink so quickly that even infinitely many of them add up to a finite total. This happens for geometric series under one condition:
$S_{\infty}$ exists when:
$|r| < 1$ that is, $-1 < r < 1$
In words: $r$ must be a proper fraction (in size).
Where the formula comes from
$S_n = \dfrac{a(1 - r^{n})}{1 - r}$
When $|r| < 1$ and $n \to \infty$, $r^{n} \to 0$. So:
$S_{\infty} = \dfrac{a(1 - 0)}{1 - r}$
Sum to infinity (in Tables):
$S_{\infty} = \dfrac{a}{1 - r}, \ \ \ |r| < 1$
Worked example — straight plug-in
Find $\dfrac{2}{3} + \dfrac{4}{9} + \dfrac{8}{27} + \dots$
$a = \dfrac{2}{3}$
$r = \dfrac{4/9}{2/3} = \dfrac{4}{9} \times \dfrac{3}{2} = \dfrac{2}{3}$
$|r| = \dfrac{2}{3} < 1$ ✓ so $S_{\infty}$ exists.
$S_{\infty} = \dfrac{a}{1 - r} = \dfrac{2/3}{1 - 2/3} = \dfrac{2/3}{1/3}$
$S_{\infty} = 2$
Worked example — negative $r$
Find $10 - 5 + \dfrac{5}{2} - \dfrac{5}{4} + \dots$
$a = 10$
$r = \dfrac{-5}{10} = -\dfrac{1}{2}$
$|r| = \dfrac{1}{2} < 1$ ✓
$S_{\infty} = \dfrac{10}{1 - (-1/2)} = \dfrac{10}{3/2}$
$S_{\infty} = \dfrac{20}{3}$
Recurring decimals as geometric series
Write $0.\dot{4}$ (that's $0.4444\dots$) as a geometric series and hence as a fraction $\dfrac{a}{b}$.
$0.\dot{4} = 0.4 + 0.04 + 0.004 + 0.0004 + \dots$
$a = 0.4, \ \ r = 0.1$
$|r| < 1$ ✓
$S_{\infty} = \dfrac{0.4}{1 - 0.1} = \dfrac{0.4}{0.9}$
$0.\dot{4} = \dfrac{4}{9}$
Mixed recurring
Write $2.\dot{5}$ in the form $\dfrac{a}{b}$.
$2.\dot{5} = 2.5555\dots = 2 + (0.5 + 0.05 + 0.005 + \dots)$
The bracket is a GP with $a = 0.5, \ r = 0.1$:
$S_{\infty} = \dfrac{0.5}{0.9} = \dfrac{5}{9}$
$2.\dot{5} = 2 + \dfrac{5}{9} = \dfrac{18}{9} + \dfrac{5}{9}$
$2.\dot{5} = \dfrac{23}{9}$
Two-digit recurring
Write $1.\dot{2}\dot{5}$ (that's $1.252525\dots$) in the form $\dfrac{a}{b}$.
$1.\dot{2}\dot{5} = 1 + 0.25 + 0.0025 + 0.000025 + \dots$
The repeating part: $a = 0.25, \ \ r = 0.01$ (each block of 2 digits is $100\times$ smaller).
$S_{\infty} = \dfrac{0.25}{1 - 0.01} = \dfrac{0.25}{0.99} = \dfrac{25}{99}$
$1.\dot{2}\dot{5} = 1 + \dfrac{25}{99} = \dfrac{99 + 25}{99}$
$1.\dot{2}\dot{5} = \dfrac{124}{99}$
Sigma form
Evaluate $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{3}{8}\right)^{n}$.
Write out the first few terms:
$\dfrac{3}{8} + \left(\dfrac{3}{8}\right)^{2} + \left(\dfrac{3}{8}\right)^{3} + \dots$
$a = \dfrac{3}{8}, \ \ r = \dfrac{3}{8}, \ \ |r| < 1$ ✓
$S_{\infty} = \dfrac{3/8}{1 - 3/8} = \dfrac{3/8}{5/8}$
$\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{3}{8}\right)^{n} = \dfrac{3}{5}$
Try this
Find $S_{\infty}$ of $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dots$
$a = \dfrac{1}{3}, \ \ r = \dfrac{1}{3}$
$S_{\infty} = \dfrac{1/3}{1 - 1/3} = \dfrac{1/3}{2/3}$
Try this
Find $S_{\infty}$ of $8 - 4 + 2 - 1 + \dots$
$a = 8, \ \ r = -\dfrac{1}{2}$
$|r| = \dfrac{1}{2} < 1$ ✓
$S_{\infty} = \dfrac{8}{1 - (-1/2)} = \dfrac{8}{3/2}$
$= 8 \times \dfrac{2}{3}$
Try this
Does $S_{\infty}$ exist for $1 + 2 + 4 + 8 + \dots$? Explain.
$a = 1, \ \ r = 2$
$|r| = 2$ -- not less than $1$.
The terms grow without bound, so no finite sum.
Try this
Write $0.\dot{7}$ (that's $0.7777\dots$) as a fraction in the form $\dfrac{a}{b}$.
$0.\dot{7} = 0.7 + 0.07 + 0.007 + \dots$
$a = 0.7, \ \ r = 0.1$
$S_{\infty} = \dfrac{0.7}{0.9}$
Try this
Write $0.\dot{3}\dot{6}$ (that's $0.363636\dots$) as a fraction in the form $\dfrac{a}{b}$.
$0.\dot{3}\dot{6} = 0.36 + 0.0036 + 0.000036 + \dots$
$a = 0.36, \ \ r = 0.01$
$S_{\infty} = \dfrac{0.36}{1 - 0.01} = \dfrac{0.36}{0.99} = \dfrac{36}{99} = \dfrac{4}{11}$
Try this
Evaluate $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{2}{5}\right)^{n}$.
Expansion: $\dfrac{2}{5} + \left(\dfrac{2}{5}\right)^{2} + \left(\dfrac{2}{5}\right)^{3} + \dots$
$a = \dfrac{2}{5}, \ \ r = \dfrac{2}{5}$
$S_{\infty} = \dfrac{2/5}{1 - 2/5} = \dfrac{2/5}{3/5}$
Part 4 done.
You now have the matched toolkit for geometric: the constant-ratio test, $T_n = ar^{n-1}$, $S_n = \dfrac{a(1 - r^{n})}{1 - r}$, sim eqs with the division trick, the $x$-question via $(T_2)^{2} = T_1 T_3$, the multiplicative symmetric labels $\dfrac{x}{y}, x, xy$, and $S_{\infty} = \dfrac{a}{1 - r}$ for $|r| < 1$ (with recurring decimals as the killer application).
That closes Sequences & Series across Parts 1–4. Compound interest and amortisation problems lean heavily on these — that's where the algebra meets the real world.