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

Recall the two facts everything below leans on, from the parent GP note: $a_{n} = a r^{n - 1}, S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{a ( r ^{n} - 1 )}{r - 1} (r \neq = 1), S_{n} = na (r = 1) .$

Here $a$ is the first term, $r$ the common ratio (multiply each term by $r$ to get the next), $n$ the term count, $a_{n}$ the $n$ -th term, $S_{n}$ the sum of the first $n$ terms.

True or false — justify

A GP can have a common ratio of

0

False — if

r = 0

, every term after the first is

0

and the ratio

a_{n + 1} / a_{n}

would need dividing by

0

, so it is not a valid GP.

A GP can have negative terms.

True — a negative

r

(or negative

a

) makes terms alternate or stay negative, e.g.

3, - 6, 12, - 24, \dots

with

r = - 2

; the ratio is still constant, so it is a genuine GP.

If all terms of a GP are equal, it is still a GP.

True — that is the case

r = 1

; the ratio is constant (

= 1

), so

a, a, a, \dots

qualifies, though its sum uses

S_{n} = na

, not the fraction formula.

Every increasing sequence of positive numbers is a GP.

False — increasing only means terms grow; a GP needs the ratio (not the difference or anything else) to be constant, e.g.

1, 2, 4, \dots

is a GP but

1, 2, 3, \dots

is not.

For a GP the sum of

n

terms always increases as

n

grows.

False — only when all terms are positive; if

r < 0

terms alternate sign, so partial sums can go up and down (e.g.

1, - 1, 1, - 1

gives sums

1, 0, 1, 0

∣ r ∣ < 1

the terms of a GP shrink toward

0

True — repeatedly multiplying by a number of size less than one drives

∣ a_{n} ∣ = ∣ a ∣ ∣ r ∣^{n - 1}

0

; this is exactly what makes the Sum of infinite GP finite.

The formula

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

works for

r = 1

False — at

r = 1

the denominator

1 - r = 0

, which is undefined; the correct value there is

S_{n} = na

found by adding

n

equal terms directly.

The two forms

\frac{a ( 1 - r ^{n} )}{1 - r}

and

\frac{a ( r ^{n} - 1 )}{r - 1}

can give different answers.

False — multiply the top and bottom of one by

- 1

and you get the other, so they are algebraically identical for every

r \neq = 1

In a GP,

a_{5} / a_{2} = r^{3}

True —

a_{5} = a r^{4}

and

a_{2} = a r

, so the ratio is

r^{4 - 1} = r^{3}

; in general

a_{p} / a_{q} = r^{p - q}

Spot the error

"For

2, 6, 18, \dots

the 4th term is

2 \cdot 3^{4} = 162

Error: the exponent should be

n - 1 = 3

, not

n

. Correct:

a_{4} = 2 \cdot 3^{3} = 54

. Always test on term 1:

a r^{0} = a

"The common ratio of

10, 7, 4, 1

r = 7/10

Error:

7/10

is only the first ratio, but

4/7 \neq = 7/10

, so the sequence is not a GP at all — it is an AP with

d = - 3

. Ratio must be constant to call it

r

"For the GP

8, 4, 2, 1

(so

a = 8, r = \frac{1}{2}

S_{4} = \frac{a ( r ^{4} - 1 )}{r - 1}

gives a negative sum, so the sum is negative."

No — with

a = 8, r = \frac{1}{2}

that form has a negative numerator and negative denominator, so they cancel to a positive value; the sum

8 + 4 + 2 + 1 = 15

is positive. Mixing forms tempts a sign slip.

r

for a GP is found the same way as

d

for an AP: subtract consecutive terms."

Error: Arithmetic Progression (AP) uses subtraction for

d

, but GP uses division —

r = a_{n + 1} / a_{n}

. Subtracting a GP gives changing differences, not a constant.

"Since

486 = 2 \cdot 3^{n - 1}

gives

3^{n - 1} = 243

, and

243

is odd,

n

must be odd."

The parity of

243

tells you nothing about

n

. Solve properly:

243 = 3^{5}

, so

n - 1 = 5

and

n = 6

(even). Compare powers, don't guess from parity.

"A GP with first term

0

and ratio

2

is fine:

0, 0, 0, \dots

Error:

a = 0

is banned because every term collapses to

0

(a degenerate sequence) and

a_{n + 1} / a_{n} = 0/0

is undefined, so no valid ratio exists.

Why questions

Why is the exponent in

a_{n}

equal to

n - 1

rather than

n

Because there are

(n - 1)

gaps (multiplications by

r

) between term 1 and term

n

; the first term is multiplied by

r

zero times, giving

r^{0} = 1

Why do we multiply

S_{n}

r

in the sum derivation rather than by some other number?

Multiplying by exactly

r

reproduces the same list of terms shifted one slot over, so subtracting the two lines cancels every interior term — a different multiplier would not line them up.

Why does the sum formula need the condition

r \neq = 1

The derivation divides both sides by

(1 - r)

; if

r = 1

that is division by zero, which is forbidden, so

r = 1

needs its own formula

S_{n} = na

Why does an infinite GP have a finite sum only when

∣ r ∣ < 1

Only then does

r^{n} \to 0

n

grows, so

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

settles at

\frac{a}{1 - r}

; if

∣ r ∣ \geq 1

the terms don't shrink and the total blows up — see Sum of infinite GP.

Why is a GP called the "discrete cousin" of an exponential function?

Both grow by repeated multiplication:

a_{n} = a r^{n - 1}

samples the curve

y = a r^{x}

at whole-number steps, so a GP is exponential growth measured in jumps.

Why do compound-interest balances form a GP?

Each period multiplies the balance by the same factor

r = 1 + \frac{i}{100}

, and constant multiplication per step is precisely the definition of a GP — this links directly to Compound interest.

Why do we take

r = a_{n + 1} / a_{n}

and not

a_{n} / a_{n + 1}

Because we move forward through the sequence: the next term equals the current one times

r

, so dividing next by current recovers that forward multiplier. Flipping it gives

1/ r

Why can we use Logarithms to find which term of a GP equals a given value?

Solving

a r^{n - 1} = k

puts the unknown

n

in an exponent; logarithms are the tool that "brings the exponent down," turning it into

n - 1 = lo g_{r} (k / a)

Edge cases

A GP with exactly one term — does the sum formula apply?

For

n = 1

S_{1} = \frac{a ( 1 - r )}{1 - r} = a

when

r \neq = 1

, which is correct; and

S_{1} = 1 \cdot a = a

when

r = 1

— both give the single term

a

What does a GP look like when

r = - 1

Terms flip sign forever:

a, - a, a, - a, \dots

; the sum is

a

for odd

n

and

0

for even

n

, and no infinite sum exists since

∣ r ∣ = 1

is not less than

1

Is a two-term sequence like

4, 12

a GP?

Yes — any two nonzero terms trivially have a ratio (

r = 3

here), so it is a (short) GP; you only run into trouble deciding consistency from three or more terms.

Can

a

be negative while

r

is positive?

Yes — e.g.

- 2, - 6, - 18

has

a = - 2, r = 3

; all terms stay negative and the ratio is still constant, so the formulas apply unchanged (watch the sign of

S_{n}

What is the geometric mean of a term-pair, and when is it real?

The middle term of a 3-term GP satisfies

b^{2} = a c

, so

b = \pm a c

(see Geometric Mean); it is real only when

a c \geq 0

, i.e. the two ends share a sign.

At the boundary

r = 1

, why is the sequence still "geometric" and not "arithmetic"?

It happens to be both: constant ratio

1

(GP) and constant difference

0

(AP). It sits exactly on the overlap of the two families, which is why it needs the special-case sum

na

Recall One-line self-test before moving on

Cover every answer above; if you can justify each in a full sentence (not "yes/no"), you have internalised the traps. The three that catch the most students: the $r = 1$ zero-denominator, the $n - 1$ vs $n$ exponent, and using subtraction to find $r$ .

Connections

Arithmetic Progression (AP) — the "subtract to find $r$ " trap comes from AP habits.
Sum of infinite GP — the $∣ r ∣ < 1$ edge cases feed directly into it.
Geometric Mean — the sign condition $a c \geq 0$ appears in the edge cases.
Exponential functions — the "discrete cousin" why-question.
Compound interest — the $r = 1 + i /100$ modelling why-question.
Logarithms — the tool for solving exponent equations.