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

Intuition How to use this page

A GP (short for geometric progression — a sequence where you multiply by a fixed number each step) is what every problem below drills. Each problem is a mini-test. Try it first with pen and paper, then open the collapsible Solution to check every step. The problems climb a ladder:

L1 Recognition — can you spot a GP and read off $a$ and $r$ ?
L2 Application — can you plug into $a_{n} = a r^{n - 1}$ and the sum formula?
L3 Analysis — can you work backwards (find $n$ , find $r$ , find $a$ )?
L4 Synthesis — can you combine GP ideas with other tools?
L5 Mastery — can you reason under pressure with a twist?

Reminders you will lean on (all built in the parent note):

nth term: $a_{n} = a r^{n - 1}$ — first term times ratio raised to the number of gaps.
sum of $n$ terms ( $r \neq = 1$ ): $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{a ( r ^{n} - 1 )}{r - 1}$ .
special case $r = 1$ : $S_{n} = na$ .
finding $r$ : divide, never subtract: $r = a_{n + 1} / a_{n}$ .

Before we start, one picture to keep in your head. The figure below draws a GP as a staircase of arrows: each arrow is a " $\times r$ " step. Notice that to reach the 4th term you cross 3 arrows — that is exactly why the nth term has exponent $n - 1$ : count the arrows (gaps), not the dots (terms). The same figure compares the GP against an AP so you can see multiplication curving away from the straight line of addition.

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

Level 1 — Recognition

Recall Solution 1.1

WHAT to do: for each list, divide each term by the one before. If that ratio is the same every time, it's a GP. WHY: the definition of a GP is a constant ratio — this is the one test.

(a) $6/2 = 3$ , $18/6 = 3$ , $54/18 = 3$ . All equal $3$ → GP with $a = 2$ , $r = 3$ . (b) $10/5 = 2$ , but $15/10 = 1.5$ . Ratios differ → not a GP. (It's an AP: you add $5$ each time — a habit trap covered below.) (c) $27/81 = \frac{1}{3}$ , $9/27 = \frac{1}{3}$ , $3/9 = \frac{1}{3}$ . All equal $\frac{1}{3}$ → GP with $a = 81$ , $r = \frac{1}{3}$ .

Recall Solution 1.2

Find $r$ by dividing: $r = \frac{- 14}{7} = - 2$ . Check: $\frac{28}{- 14} = - 2$ ✓. WHY the sign flips: a negative ratio multiplies the sign each step, so terms alternate positive/negative — this is a perfectly valid GP. Next term: $- 56 \times (- 2) = 112$ .

Level 2 — Application

Recall Solution 2.1

WHAT: use $a_{n} = a r^{n - 1}$ with $n = 10$ . WHY exponent $9$ : there are $n - 1 = 10 - 1 = 9$ gaps (multiplications) between term 1 and term 10 — count the arrows, not the terms (exactly the picture in figure s01). $a_{10} = 3 \cdot 2^{9} = 3 \cdot 512 = 1536 .$

Recall Solution 2.2

Identify: $a = 1$ , $r = 3/1 = 3$ , $n = 8$ . Pick the form: since $r > 1$ , use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ so both top and bottom stay positive (no sign slips). $S_{8} = \frac{1 ( 3 ^{8} - 1 )}{3 - 1} = \frac{6561 - 1}{2} = \frac{6560}{2} = 3280 .$

Recall Solution 2.3

Identify: $a = 16$ , $r = 8/16 = \frac{1}{2}$ , $n = 5$ . Pick the form: since $r < 1$ , use $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ so $1 - r > 0$ . $S_{5} = \frac{16 ( 1 - ( \frac{1}{2} ) ^{5} )}{1 - \frac{1}{2}} = \frac{16 ( 1 - \frac{1}{32} )}{\frac{1}{2}} = 32 \cdot \frac{31}{32} = 31 .$ Verify by hand: $16 + 8 + 4 + 2 + 1 = 31$ ✓.

Level 3 — Analysis

Recall Solution 3.1

WHAT: we know $a_{n}$ and want $n$ — work backwards. $a = 3$ , $r = 2$ . Set $a r^{n - 1} = 768$ : $3 \cdot 2^{n - 1} = 768 \Rightarrow 2^{n - 1} = 256.$ WHY next: recognise $256 = 2^{8}$ , so $n - 1 = 8 \Rightarrow n = 9$ . (If you can't spot the power, take a logarithm: $n - 1 = lo g_{2} 256 = 8$ .)

Recall Solution 3.2

WHAT: write both facts with $a_{n} = a r^{n - 1}$ , then divide to kill $a$ . $a_{3} = a r^{2} = 20, a_{6} = a r^{5} = 160.$ WHY divide: dividing the equations cancels $a$ and leaves only $r$ : $\frac{a r ^{5}}{a r ^{2}} = \frac{160}{20} \Rightarrow r^{3} = 8 \Rightarrow r = 2.$ Back-substitute: $a r^{2} = 20 \Rightarrow a \cdot 4 = 20 \Rightarrow a = 5$ , and $r = 2$ .

Recall Solution 3.3

Set up: $a = 2$ , $r = 2$ , use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{2 ( 2 ^{n} - 1 )}{1} = 2 (2^{n} - 1)$ . Solve the inequality $S_{n} \geq 2046$ : $2 (2^{n} - 1) \geq 2046 \Rightarrow 2^{n} - 1 \geq 1023 \Rightarrow 2^{n} \geq 1024 = 2^{10} .$ So $n \geq 10$ . The 10th partial sum is the first to reach it: $S_{10} = 2 (1024 - 1) = 2046$ exactly. Answer: $n = 10$ .

Level 4 — Synthesis

Recall Solution 4.1

Connect to Compound interest: each year multiplies the amount by $r = 1 + \frac{10}{100} = 1.1$ . Careful with indexing (avoid an off-by-one trap): the principal $₹1000$ is the value at time zero — call it $A_{0} = 1000$ . After $k$ full years the amount is $A_{0} r^{k}$ , because $k$ years means $k$ multiplications by $r$ . So "after 3 years" uses the exponent $3$ (three multiplications), not $n - 1$ . This matches $a_{n} = a r^{n - 1}$ only if we relabel the year-1 amount as the first term $a$ : then $a = A_{0} r = 1100$ and the year- $m$ amount is $a r^{m - 1}$ . Both viewpoints agree; just be consistent about which slot you call "term 1". Amount after 3 years: $A_{0} r^{3} = 1000 \times (1.1)^{3} = 1000 \times 1.331 = ₹ 1331$ . Year-end amounts: year 1 = $1100$ , year 2 = $1210$ , year 3 = $1331$ — a GP with first term $a = 1100$ , $r = 1.1$ , $n = 3$ . $Total = \frac{1100 ( 1. 1 ^{3} - 1 )}{1.1 - 1} = \frac{1100 ( 1.331 - 1 )}{0.1} = \frac{1100 \times 0.331}{0.1} = ₹ 3641 .$

Recall Solution 4.2

Clever labelling: write the three terms as $\frac{b}{r}, b, b r$ (using the middle as the Geometric Mean). WHY: the product then collapses beautifully. Product: $\frac{b}{r} \cdot b \cdot b r = b^{3} = 64 \Rightarrow b = 4$ . Sum: $\frac{4}{r} + 4 + 4 r = 14 \Rightarrow \frac{4}{r} + 4 r = 10$ . Multiply by $r$ : $4 r^{2} - 10 r + 4 = 0 \Rightarrow 2 r^{2} - 5 r + 2 = 0 \Rightarrow (2 r - 1) (r - 2) = 0$ . So $r = 2$ or $r = \frac{1}{2}$ . Both give the same set of numbers: $2, 4, 8$ .

Level 5 — Mastery

Recall Solution 5.1

What goes wrong: with $r = 1$ , the denominator $1 - r = 0$ , so the formula is $\frac{7 ( 1 - 1 )}{0} = \frac{0}{0}$ — undefined (division by zero). The formula was derived by dividing by $1 - r$ , which is only legal when $r \neq = 1$ . Correct reasoning: if $r = 1$ every term equals $a$ , so the sequence is $7, 7, 7, 7, 7$ and $S_{5} = 5 \times 7 = 35 (i.e. S_{n} = na) .$

Recall Solution 5.2

Find $r$ : first two terms are $a + a r = a (1 + r) = 12 (1 + r) = 16$ , so $1 + r = \frac{16}{12} = \frac{4}{3} \Rightarrow r = \frac{1}{3}$ . Since $∣ \frac{1}{3} ∣ < 1$ , the infinite sum exists. Connect to Sum of infinite GP: because $∣ r ∣ < 1$ , $r^{n} \to 0$ as $n \to \infty$ , so $S_{\infty} = \frac{a}{1 - r}$ . $S_{\infty} = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 12 \times \frac{3}{2} = 18 .$

Recall Solution 5.3

Set up the sum: with $a = 2$ , $S_{4} = a (1 + r + r^{2} + r^{3}) = 2 (1 + r + r^{2} + r^{3}) = 30 \Rightarrow 1 + r + r^{2} + r^{3} = 15.$ Solve by factor-hunting: try small whole numbers. $r = 2$ : $1 + 2 + 4 + 8 = 15$ ✓. So $r = 2$ . WHY guessing is legitimate here: the terms $1, r, r^{2}, r^{3}$ grow fast, so only a couple of integer candidates can land on $15$ — testing $r = 2$ is quicker than solving the cubic by formula. Verify: the GP is $2, 4, 8, 16$ , and $2 + 4 + 8 + 16 = 30$ ✓. Using the formula: $S_{4} = \frac{2 ( 2 ^{4} - 1 )}{2 - 1} = 2 \times 15 = 30$ ✓.

Connections

Parent note (Hinglish) — the derivations these exercises drill.
Sum of infinite GP — Exercise 5.2 lives here.
Geometric Mean — the $\frac{b}{r}, b, b r$ trick in Exercise 4.2.
Compound interest — Exercise 4.1's real-world GP.
Logarithms — the escape hatch for "find $n$ " when powers don't factor nicely.
Arithmetic Progression (AP) — the source of the "subtract to get the step" trap.