Geometric progression (GP) — nth term, sum of n terms — derivations: Worked examples

Everything below uses only two tools from the parent: $a_{n} = a r^{n - 1}, S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{a ( r ^{n} - 1 )}{r - 1} (r \neq = 1) .$ Here $a$ = first term, $r$ = common ratio (the number you multiply by each step), $n$ = how many terms.

The scenario matrix

Every GP problem lives in one of these cells. The last column names the example that kills it.

#	Case class	What's special / where it bites	Example
A	$r > 1$ (growth)	terms explode; use $\frac{a ( r ^{n} - 1 )}{r - 1}$ to stay positive	Ex 1
B	$0 < r < 1$ (decay)	terms shrink; use $\frac{a ( 1 - r ^{n} )}{1 - r}$	Ex 2
C	$r < 0$ (alternating sign)	terms flip $+, -, +, -$ ; sign of $r^{n - 1}$ matters	Ex 3
D	$r = 1$ (degenerate)	denominator $1 - r = 0$ → formula illegal, use $S_{n} = na$	Ex 4
E	find $n$ (unknown count)	rearrange nth-term, use Logarithms	Ex 5
F	find $r$ from two terms	divide, take a root — watch $\pm$	Ex 6
G	real world: Compound interest	$r = 1 + \frac{i}{100}$ , a word problem	Ex 7
H	limiting case $∣ r ∣ < 1$ , $n \to \infty$	connects to Sum of infinite GP	Ex 8

Figure — the four ratio-behaviours side by side, so you can see which cell you're in before touching algebra:

Figure — Geometric progression (GP) — nth term, sum of n terms — derivations

Example 1 — Cell A: pure growth, $r > 1$

Worked example Ex 1

A GP starts $2, 6, 18, \dots$ Find (a) the 8th term and (b) the sum of the first 8 terms.

Forecast: the terms triple each time, so the 8th term is big — several thousand. Guess before reading on.

Step 1. Identify $a$ and $r$ . $a = 2$ ; $r = \frac{6}{2} = 3$ . Why this step? $r$ is defined by dividing consecutive terms (never subtracting — that's the AP habit).

Step 2. 8th term: $a_{8} = a r^{8 - 1} = 2 \cdot 3^{7}$ . Why $7$ ? There are $8 - 1 = 7$ multiplication-arrows between term 1 and term 8. Count arrows, not terms. $a_{8} = 2 \cdot 2187 = 4374.$

Step 3. Sum. Since $r = 3 > 1$ , use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ so both parts stay positive. Why this form? With $r > 1$ , $r^{n} - 1 > 0$ and $r - 1 > 0$ — no sign gymnastics. $S_{8} = \frac{2 ( 3 ^{8} - 1 )}{3 - 1} = \frac{2 ( 6561 - 1 )}{2} = 6560.$

Verify: ratio check $6 \div 2 = 3$ , $18 \div 6 = 3$ ✓. And $a_{8} = 4374$ should sit below $S_{8} = 6560$ (the last term is always less than the running total) ✓.

Example 2 — Cell B: decay, $0 < r < 1$

Example 3 — Cell C: negative ratio, alternating signs

Worked example Ex 3

A GP is $3, - 6, 12, - 24, \dots$ Find (a) the 6th term and (b) the sum of the first 6 terms.

Forecast: signs go $+, -, +, -, +, -$ . So the 6th term is negative. And the sum? Positives and negatives partly cancel, so it should be small in size.

Step 1. $a = 3$ ; $r = \frac{- 6}{3} = - 2$ . Why negative? Dividing a term by the previous one gives $- 2$ ; the minus sign is what flips each term's sign.

Step 2. 6th term: $a_{6} = 3 \cdot (- 2)^{6 - 1} = 3 \cdot (- 2)^{5}$ . Why track the exponent's parity? $(- 2)^{5} = - 32$ (odd power keeps the minus). So $a_{6} = 3 \cdot (- 32) = - 96,$ confirming our forecast that even-position terms are negative.

Step 3. Sum. Use $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ ; here $r^{6} = (- 2)^{6} = + 64$ (even power → positive). Why check parity again? Getting $r^{n}$ 's sign wrong is the #1 error with negative $r$ . $S_{6} = \frac{3 ( 1 - 64 )}{1 - ( - 2 )} = \frac{3 ( - 63 )}{3} = - 63.$

Verify: direct add $3 - 6 + 12 - 24 + 48 - 96$ . Group: $(3 - 6) + (12 - 24) + (48 - 96) = - 3 - 12 - 48 = - 63$ ✓. Small-ish and negative, as forecast.

Example 4 — Cell D: the degenerate $r = 1$

Example 5 — Cell E: find the unknown count $n$

Worked example Ex 5

In the GP $5, 15, 45, \dots$ , which term equals $10935$ ?

Forecast: terms triple each step. Rough size: $5 \to 15 \to 45 \to 135 \to 405 \to 1215 \to 3645 \to 10935$ . Count them mentally — around the 8th term.

Step 1. $a = 5$ , $r = 3$ . Set up the nth-term equation: $a r^{n - 1} = 10935 \Rightarrow 5 \cdot 3^{n - 1} = 10935.$ Why this equation? We know the term's value, not its position — so we solve for $n$ .

Step 2. Isolate the power: $3^{n - 1} = \frac{10935}{5} = 2187$ . Why divide first? To strip $a$ and leave a clean power of $r$ .

Step 3. Recognise $2187 = 3^{7}$ (since $3^{7} = 2187$ ). Match exponents: $n - 1 = 7 \Rightarrow n = 8.$ Why can we match exponents? If bases are equal and positive, the exponents must be equal. (If $2187$ weren't a neat power, we'd take a logarithm: $n - 1 = lo g_{3} 2187$ .)

Verify: $a_{8} = 5 \cdot 3^{7} = 5 \cdot 2187 = 10935$ ✓. Matches the forecast (the 8th term).

Example 6 — Cell F: find $r$ from two terms (the $\pm$ subtlety)

Worked example Ex 6

In a GP the 2nd term is $6$ and the 4th term is $54$ . Find the possible values of $r$ and of the first term $a$ .

Forecast: going from term 2 to term 4 multiplies by $r$ twice, and the value went $6 \to 54$ , i.e. $\times 9$ . So $r^{2} = 9$ — and a square opens the door to two answers.

Step 1. Write both terms: $a_{2} = a r = 6$ and $a_{4} = a r^{3} = 54$ . Why both? Two unknowns ( $a, r$ ) need two equations.

Step 2. Divide the equations to kill $a$ : $\frac{a r ^{3}}{a r} = \frac{54}{6} \Rightarrow r^{2} = 9.$ Why divide? It removes $a$ instantly, leaving one equation in $r$ .

Step 3. Take the square root — both signs: $r = + 3 or r = - 3.$ Why keep both? A square root has two roots; a real GP can genuinely have a negative ratio (Cell C). Never drop the minus without a reason.

Step 4. Back-substitute into $a r = 6$ :

If $r = 3$ : $a = 6/3 = 2$ → GP $2, 6, 18, 54, \dots$
If $r = - 3$ : $a = 6/ (- 3) = - 2$ → GP $- 2, 6, - 18, 54, \dots$

Verify: for $r = 3$ : $a_{4} = 2 \cdot 3^{3} = 54$ ✓. For $r = - 3$ : $a_{4} = - 2 \cdot (- 3)^{3} = - 2 \cdot (- 27) = 54$ ✓. Both are legitimate — the problem has two answers, and forgetting one loses marks.

Example 7 — Cell G: real-world compound interest

Worked example Ex 7

₹10000 is invested at $8%$ per year, compounded annually. What is the amount after $5$ years, and how much interest was earned?

Forecast: $8%$ growth for 5 years is a bit more than $40%$ total (because interest itself earns interest), so expect around ₹14000–15000.

Step 1. Model as a GP of yearly amounts. Each year the amount is multiplied by $1 + \frac{8}{100} = 1.08$ . So Compound interest gives the ratio $r = 1 + \frac{i}{100} = 1.08, a_{1} = 10000 \times 1.08 (end of year 1) .$ Why is this a GP? Multiplying by the same factor each year is exactly the GP rule — repeated multiplication.

Step 2. The amount after $t$ years is $A = P (1 + \frac{i}{100})^{t} = 10000 \cdot (1.08)^{5} .$ Why exponent $t$ (not $t - 1$ )? Here $P$ is the amount before any growth (year 0); it gets multiplied once per year, $5$ times. Contrast the nth-term convention where the first term is already $a$ . Always ask: is my starting value "before" or "after" the first multiplication?

Step 3. Compute $(1.08)^{5} = 1.4693280768$ , so $A = 10000 \cdot 1.4693280768 = 14693.28 (₹, to the paisa) .$

Step 4. Interest = amount − principal: $14693.28 - 10000 = 4693.28 ₹ .$

Verify: simple interest would be $5 \times 8% = 40%$ = ₹4000; compound gives more (₹4693.28) because earlier interest compounds — consistent with the forecast, and with the "multiply, don't add" spirit of Exponential functions.

Example 8 — Cell H: limiting behaviour, $n \to \infty$

Worked example Ex 8

A bouncing-ball height sequence is $10, 5, 2.5, 1.25, \dots$ metres. (a) Find $S_{6}$ . (b) What total does the sum approach as the number of terms grows without bound?

Forecast: each height halves, so after many bounces the total settles near a fixed number — the tail contributes almost nothing. Guess the ceiling.

Step 1. $a = 10$ , $r = \frac{1}{2}$ (Cell B). Six-term sum: $S_{6} = \frac{10 ( 1 - ( \frac{1}{2} ) ^{6} )}{1 - \frac{1}{2}} = \frac{10 ( 1 - \frac{1}{64} )}{\frac{1}{2}} = 20 \cdot \frac{63}{64} = \frac{1260}{64} = 19.6875.$ Why the $\frac{a ( 1 - r ^{n} )}{1 - r}$ form? $r < 1$ keeps everything positive.

Step 2. Now let $n$ grow. Look at $r^{n} = (\frac{1}{2})^{n}$ : as $n$ gets huge, halving again and again drives it toward $0$ . Why does $r^{n} \to 0$ ? Because $∣ r ∣ < 1$ — repeatedly multiplying by a number smaller than 1 shrinks it without limit. (If $∣ r ∣ \geq 1$ this would not happen and no finite total exists.)

Step 3. Replace $r^{n}$ by $0$ in the formula — this is exactly the Sum of infinite GP result: $S_{\infty} = \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r} = \frac{10}{1 - \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 20 m .$

Verify: $S_{6} = 19.6875$ is already within $0.32$ m of the ceiling $20$ ; each extra bounce closes half the remaining gap ✓. The limit exists precisely because $∣ r ∣ < 1$ — the boundary condition of the whole idea.

Recall Which cell am I in? (self-quiz)

A GP has $r = - \frac{1}{3}$ ; is the 4th term positive or negative? ::: Negative — $(- \frac{1}{3})^{3}$ has an odd power, so the sign flips. Your sum formula gives $\frac{0}{0}$ . What happened and what do you do? ::: $r = 1$ (degenerate cell); use $S_{n} = na$ instead. $r^{2} = 16$ from two terms — how many ratios are possible? ::: Two: $r = + 4$ and $r = - 4$ (keep both unless the problem forbids one). When does adding infinitely many GP terms give a finite total? ::: Only when $∣ r ∣ < 1$ , giving $S_{\infty} = \frac{a}{1 - r}$ . For $r > 1$ which sum form avoids sign slips? ::: $\frac{a ( r ^{n} - 1 )}{r - 1}$ (both parts positive).

Connections

Parent (Hinglish) — the formula derivations these examples exercise.
Arithmetic Progression (AP) — contrast: $r$ by division here, $d$ by subtraction there.
Sum of infinite GP — Ex 8's limit $\frac{a}{1 - r}$ .
Geometric Mean — the middle-term relation behind two-term problems like Ex 6.
Exponential functions — Ex 7's continuous cousin.
Compound interest — Ex 7's model, $r = 1 + \frac{i}{100}$ .
Logarithms — Ex 5's tool when $r^{n - 1} = k$ isn't a clean power.