Finite geometric sums | Year 11

01 / LEARN

Understand the idea.

Subtracting a shifted sum

To sum $S_{n} = a + a r + \dots + a r^{n - 1}$ , compare it with $r S_{n}$ . Subtracting leaves only the first term and the shifted final term: $(1 - r) S_{n} = a - a r^{n}$ . This explains both the numerator and denominator of the geometric-sum formula.

Division requires $r \neq 1$ . For $r = 1$ , every term is $a$ , giving $S_{n} = n a$ . A finite geometric sum is valid for any real ratio; it does not require $∣ r ∣ < 1$ .

S_{n} = {\begin{cases} \frac{a (1 - r^{n})}{1 - r}, & r \neq 1 \\ n a, & r = 1 \end{cases}

Identify the actual powers

Count the number of included terms before substituting. A negative ratio needs brackets when raised to a power, since ${(- 2)}^{6}$ and $- 2^{6}$ have opposite signs. Either equivalent fraction form works if its numerator and denominator signs match.

For repeated deposits, list how long each payment earns interest. Immediately after an end-of-period payment, that newest payment has exponent zero. Older payments have successively larger exponents; the balance is the sum of these separate contributions.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An alternating finite sum

Find the sum of the first six terms of $3, - 6, 12, \dots$ .

Worked solution

1
Read the geometric parameters.
$a = 3, r = - 2, n = 6$
The sum contains six terms and each term is multiplied by negative two.
2
Substitute with the negative ratio bracketed.
$S_{6} = \frac{3 [1 - {(- 2)}^{6}]}{1 - (- 2)}$
The finite-sum formula applies because the ratio is not one.
3
Evaluate the power and simplify.
$S_{6} = \frac{3 (1 - 64)}{3} = - 63$
The positive denominator does not cancel the negative sign of the numerator.

Answer

$- 63$ .

EXAMPLE 02

End-of-year deposits

AUD 200 is paid at the end of each year for four years, with 5% annual interest. Find the balance immediately after payment 4.

Worked solution

1
List the accumulated contributions.
$B = 200 {(1.05)}^{3} + 200 {(1.05)}^{2} + 200 (1.05) + 200$
The first payment earns three years of interest; the fourth earns none yet.
2
Order the sum by increasing powers.
$B = 200 [1 + 1.05 + {(1.05)}^{2} + {(1.05)}^{3}]$
Addition order can change without changing the balance.
3
Apply the four-term sum.
$B = 200 \frac{{(1.05)}^{4} - 1}{1.05 - 1} = 862.025$
The geometric ratio is the annual growth factor, with four deposits.
4
Round the final balance to cents.
$B \approx AUD 862.03$
Rounding only the final result preserves the supplied annual-interest model.

Answer

$862.03$ to the nearest cent.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find $1 + 3 + 9 + 27 + 81$ .

Show answer & reasoning

$121$ .

There are five terms with first term one and ratio three, so $S_{5} = (3^{5} - 1) / (3 - 1) = 242 / 2 = 121$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find the sum of eight terms of $5, 10, 20, \dots$ .
2 marks
Find $16 + 8 + 4 + 2 + 1 + \frac{1}{2}$ .
2 marks
Find the seven-term sum when $a = 4, r = 1$ .
2 marks
Find $2 - 6 + 18 - 54 + 162$ .
2 marks
For $a = 7, r = 2$ , find the least $n$ with $S_{n} > 1000$ .
2 marks
Find the balance immediately after three end-of-year AUD 500 deposits earning 4% yearly.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the sum of eight terms of $5, 10, 20, \dots$ .
2 marks
$5 (2^{8} - 1) = 1275$ .
$1275$ .
Find $16 + 8 + 4 + 2 + 1 + \frac{1}{2}$ .
2 marks
$\frac{16 (1 - 2^{- 6})}{1 - \frac{1}{2}} = \frac{63}{2}$ .
$\frac{63}{2}$ .
Find the seven-term sum when $a = 4, r = 1$ .
2 marks
Seven equal terms total $7 (4) = 28$ .
$28$ .
Find $2 - 6 + 18 - 54 + 162$ .
2 marks
$\frac{2 [1 - {(- 3)}^{5}]}{4} = 122$ .
$122$ .
For $a = 7, r = 2$ , find the least $n$ with $S_{n} > 1000$ .
2 marks
$S_{7} = 889$ and $S_{8} = 1785$ .
$8$ .
Find the balance immediately after three end-of-year AUD 500 deposits earning 4% yearly.
3 marks
$500 [{(1.04)}^{2} + 1.04 + 1] = 1560.8$ .
$1560.80$ .

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Subtracting a shifted sum

Identify the actual powers

See the method in action.

An alternating finite sum

Read the geometric parameters.

Substitute with the negative ratio bracketed.

Evaluate the power and simplify.

End-of-year deposits

List the accumulated contributions.

Order the sum by increasing powers.

Apply the four-term sum.

Round the final balance to cents.

Try it yourself.

Check your understanding

Keep practising