Geometric Series Formula: Savings Plans, Loan Repayments and the Sum to Infinity

In 30 seconds

An arithmetic series adds a fixed amount each time; a geometric series multiplies by a fixed ratio each time.
The geometric series formula $ S_n = \dfrac{a(r^n-1)}{r-1} $ adds $ n $ terms in one step, however many there are.
Money problems are geometric series in disguise: savings with growing deposits, compound interest on regular payments, and loan repayments.
If $ |r| < 1 $, the terms shrink fast enough for the series to have a finite sum to infinity, $ S_\infty = \dfrac{a}{1-r} $. That is how a perpetuity is priced.
Over long periods, geometric growth always overtakes arithmetic growth.

The case: two savings plans for a flat deposit

A new graduate wants to save for the deposit on a flat. In the first year she can put aside $3,000, and she expects to save more each year as her pay rises. She is weighing two plans (the figures are illustrative):

Plan A: raise her annual saving by a fixed $200 each year: $3,000, $3,200, $3,400, and so on.
Plan B: raise her annual saving by 6% each year: $3,000, $3,180, $3,370.80, and so on.

Plan A looks better at first, since $200 is more than 6% of $3,000. But which plan saves more in total over 10 years? And over 20? Adding up 20 numbers by hand is slow and easy to get wrong. The formulas for arithmetic and geometric series give the answer in a few lines.

Later in the post the same tools answer two other money questions: how a loan repayment is calculated, and how much a fund must hold to pay a scholarship for ever.

The concept: adding versus multiplying

A sequence is a list of numbers that follows a rule; a series is what you get when you add them up. Plan A is an arithmetic sequence: each term is the previous one plus a common difference $ d = 200 $. Plan B is a geometric sequence: each term is the previous one times a common ratio $ r = 1.06 $.

The arithmetic sum has a neat trick, often told in a story about the young Gauss. Write the series forwards and backwards and add the two rows. Each pair (first + last, second + second-last, and so on) has the same total, so the sum is the number of terms times the average of the first and last terms.

The geometric sum has a different trick. Multiply the whole series by $ r $. Almost every term reappears, shifted one place along. Subtract the two versions and everything cancels except the first and last pieces. That single cancellation is the reason the geometric series formula exists, and it works for any number of terms^[1].

Key termGeometric series: the sum $ a + ar + ar^2 + \dots + ar^{n-1} $, where $ a $ is the first term and $ r $ is the common ratio. Each term is $ r $ times the one before it.

The formula

For an arithmetic series with first term $ a $, common difference $ d $ and $ n $ terms:

\[ S_n = \frac{n}{2}(a + \ell) \] \[ \ell = a + (n-1)d \]

For a geometric series with first term $ a $, common ratio $ r \ne 1 $ and $ n $ terms:

\[ S_n = \frac{a(r^n-1)}{r-1} \]

and, when $ -1 < r < 1 $, the sum to infinity:

\[ S_\infty = \frac{a}{1-r} \]

Here $ S_n $ is the sum of the first $ n $ terms, $ \ell $ is the last ($ n $th) term of the arithmetic series, and the $ n $th term of a geometric series is $ ar^{n-1} $. Substituting for $ \ell $ gives the familiar form $ S_n = \frac{n}{2}\big(2a + (n-1)d\big) $. The geometric formula can also be written $ \dfrac{a(1-r^n)}{1-r} $, which is tidier when $ r < 1 $. Both forms are the same.

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Why this formula fits the case

Plan A grows by adding, so its yearly savings form an arithmetic sequence and its total is an arithmetic series. Plan B grows by a percentage, which means multiplying by 1.06 each year, so its total is a geometric series. Matching the formula to the rule of growth is the whole skill: fixed amount means arithmetic, fixed percentage means geometric.

The formulas are the wrong tool when the pattern is neither. If her pay rise varies from year to year, there is no common difference or ratio, and you have to add the actual figures (a spreadsheet does this well). The sum to infinity is also a trap if $ |r| \ge 1 $: Plan B’s ratio is 1.06, so its terms grow without limit, and $ \dfrac{a}{1-r} $ would give a meaningless negative number.

Setting it up

	Plan A (arithmetic)	Plan B (geometric)
First term	$ a = 3000 $	$ a = 3000 $
Rule	$ d = 200 $	$ r = 1.06 $
Number of terms	$ n = 10 $ or $ 20 $	$ n = 10 $ or $ 20 $
Year $ n $ saving	$ 3000 + 200(n-1) $	$ 3000(1.06)^{n-1} $

All amounts are in dollars. For now we ignore interest, so we are comparing the cash she puts in. (Adding interest makes Plan B look even better, because its larger later deposits are the ones still growing.)

Solving it

Worked example 1 (the two plans)

Find the total saved under each plan after 10 years and after 20 years.

Step 1Plan A, 10 years: $ S_{10} = \dfrac{10}{2}\big(2(3000) + 9(200)\big) $, which is $ 5(6000 + 1800) $.

Step 2So $ S_{10} = 5 \times 7800 = 39000 $ dollars.

Step 3Plan B, 10 years: $ S_{10} = \dfrac{3000(1.06^{10}-1)}{1.06-1} $, which is $ \dfrac{3000(1.790848-1)}{0.06} $.

Step 4So $ S_{10} = 50000 \times 0.790848 \approx 39542.38 $ dollars.

Step 5Plan A, 20 years: $ S_{20} = 10\big(6000 + 19(200)\big) $, which is $ 10 \times 9800 = 98000 $ dollars.

Step 6Plan B, 20 years: $ S_{20} = 50000(1.06^{20}-1) $, which is $ 50000 \times 2.207135 \approx 110356.77 $ dollars.

After 10 years Plan B is ahead by about $542. After 20 years it is ahead by about $12,357.

Total saved: Plan A (+$200 a year) vs Plan B (+6% a year) $

Plan A: arithmeticPlan B: geometric

Figure 1. Plan A adds a fixed $200 each year, so its total grows like a quadratic. Plan B grows each deposit by 6%, so its total grows exponentially. The two are almost level for seven years; Plan B moves ahead in year 8 and the gap widens from then on.

The second case shows that loans are geometric series too. A lender charges interest on the balance each month, and each repayment reduces it. Setting the final balance to zero gives an equation with a geometric series inside it.

Worked example 2 (a loan repayment)

A $12,000 car loan charges 1% interest per month and is repaid by 24 equal monthly payments of $ M $, the first one month after the loan. Find $ M $ and the total interest paid.

Step 1If no payments were made, the debt after 24 months would be $ 12000(1.01)^{24} $.

Step 2Each payment also grows until month 24. The first grows for 23 months and the last for none, so together they are worth $ M(1 + 1.01 + 1.01^2 + \dots + 1.01^{23}) $.

Step 3That bracket is a geometric series with $ a = 1 $, $ r = 1.01 $, $ n = 24 $: it equals $ \dfrac{1.01^{24}-1}{0.01} $.

Step 4The loan is cleared when the two match: $ 12000(1.01)^{24} = M \cdot \dfrac{1.01^{24}-1}{0.01} $.

Step 5With $ 1.01^{24} \approx 1.269735 $: $ M = \dfrac{12000 \times 1.269735 \times 0.01}{0.269735} $, so $ M \approx 564.88 $ dollars.

Step 6Total repaid: $ 24 \times 564.88 = 13557.12 $, so interest is about $ 13557.12-12000 = 1557.12 $ dollars.

The monthly payment is $564.88, and the loan costs about $1,557 in interest.

Worked example 3 (sum to infinity: a scholarship for ever)

A college wants to fund a $5,000 scholarship every year, for ever, with the first award in one year’s time. Money in its fund earns 4% a year. How much must it invest today?

Step 1An award paid in $ k $ years needs $ \dfrac{5000}{1.04^k} $ invested today.

Step 2So the fund needed is $ \dfrac{5000}{1.04} + \dfrac{5000}{1.04^2} + \dots $, a geometric series.

Step 3Here $ a = \dfrac{5000}{1.04} $ and $ r = \dfrac{1}{1.04} $. Since $ |r| < 1 $, the sum to infinity exists.

Step 4$ S_\infty = \dfrac{5000/1.04}{1-1/1.04} $. Multiply top and bottom by 1.04 to get $ \dfrac{5000}{1.04-1} $, which is $ \dfrac{5000}{0.04} $.

Step 5So the fund needed is $ 125000 $ dollars.

A fund of $125,000 earns exactly $5,000 a year at 4%, so it can pay the award each year without ever shrinking. That is a neat check on the answer.

Practise APs, GPs, loans and sums to infinity with 1,565 questions, every one fully worked.Get 1,565 questions · $9

What the answer tells us

In the savings case, Plan A’s yearly saving is bigger only until year 5. In year 6, Plan B saves $ 3000(1.06)^5 \approx 4014.68 $, against Plan A’s $4,000, and from then on the gap grows each year. The running totals take longer to swap, because Plan A built up an early lead: Plan A is still ahead after 7 years ($25,200 against about $25,182), and Plan B moves ahead in year 8. Over a career, the lesson is that a percentage rise beats a fixed rise, as long as you give it enough time.

The loan answer is reasonable too. Spread evenly with no interest, $12,000 over 24 months would be $500 a month. With 1% a month the payment is higher, at $564.88, and the interest of about $1,557 is about 13% of the amount borrowed. That fits two years at 12% a year on a balance that falls from $12,000 to zero.

The perpetuity shows how sensitive long-run sums are to the rate. At 5% the fund needed falls to $ 5000/0.05 = 100000 $ dollars; at 2% it rises to $ 5000/0.02 = 250000 $ dollars. Halving the rate doubles the fund, because the formula $ a/(1-r) $ is very sensitive when $ r $ is close to 1. The same formula sits behind the economist’s spending multiplier and the present value of a bond that never matures^[2].

Common mistakeUsing the 20th term instead of the 20-term sum, or the wrong power. The $ n $th term of a GP is $ ar^{n-1} $, but the sum uses $ r^n $. In Plan B, the year-20 saving is $ 3000(1.06)^{19} \approx 9076.80 $, while the 20-year total is $ 50000(1.06^{20}-1) $. Mixing these up is one of the most common errors in exam scripts.

Common mistakeUsing $ S_\infty $ when $ |r| \ge 1 $. The sum to infinity only exists when the terms shrink. Always state “since $ |r| < 1 $, $ S_\infty $ exists” before using the formula; examiners often give a mark for it.

Exam tipCambridge 9709 Pure Mathematics 1 (topic 1.6) covers arithmetic and geometric progressions, the condition for convergence and the sum to infinity^[3]. IB Mathematics: analysis and approaches SL 1.2, 1.3 and 1.8 cover the same formulas, and SL 1.4 sets them in financial contexts such as loans and compound interest^[4]. In AP Calculus BC, geometric series return in Unit 10 (topic 10.2) as the first series whose convergence you can test exactly^[5].

The Mathematics Made Simple textbook develops these formulas in Chapter 6 and returns to infinite series in Chapter 20, while the Mathematics Made Simple practice questions book gives you plenty of money-based problems with full solutions.

Get both Mathematics Made Simple books for $16: learn series, then practise them to exam standard.Both books · $16

Practice questions

EasyQ1. A salary starts at $32,000 and rises by $1,500 each year. Find the salary in year 8 and the total earned over the first 8 years. Show worked answer

This is an AP with $ a = 32000 $ and $ d = 1500 $.
Year 8: $ 32000 + 7(1500) = 42500 $ dollars.
Total: $ S_8 = \dfrac{8}{2}\big(2(32000) + 7(1500)\big) $, which is $ 4(74500) = 298000 $ dollars.

EasyQ2. Find the sum of the first 7 terms of the geometric series $ 5 + 15 + 45 + \dots $. Show worked answer

$ a = 5 $ and $ r = 3 $.
$ S_7 = \dfrac{5(3^7-1)}{3-1} = \dfrac{5(2187-1)}{2} $, which is $ \dfrac{5 \times 2186}{2} = 5465 $.

MediumQ3. A ball is dropped from a height of 10 m. Each time it lands, it bounces back to 60% of the height it fell from. Find the total distance it travels before it comes to rest. Show worked answer

It falls 10 m first. After that, each bounce goes up and comes down the same height: 6 m, 3.6 m, 2.16 m, and so on.
Those heights form a GP with $ a = 6 $, $ r = 0.6 $; since $ |r| < 1 $, $ S_\infty = \dfrac{6}{1-0.6} = 15 $ m.
Total distance: $ 10 + 2 \times 15 = 40 $ m.

MediumQ4. Someone saves $100 in the first month and 4% more each month after that. After how many whole months does the total first exceed $5,000? Show worked answer

Need $ \dfrac{100(1.04^n-1)}{0.04} > 5000 $, so $ 2500(1.04^n-1) > 5000 $.
That gives $ 1.04^n > 3 $, so $ n > \dfrac{\ln 3}{\ln 1.04} \approx 28.01 $.
So $ n = 29 $ months. Check: $ S_{28} \approx 4996.76 $ and $ S_{29} \approx 5296.63 $.

ExamQ5. A student borrows $20,000 at 0.5% interest per month. She repays $ M $ dollars at the end of each month. (a) Show that after 2 payments she owes $ 20000(1.005)^2-M(1.005 + 1) $. [2] (b) Show that after $ n $ payments she owes $ 20000(1.005)^n-200M(1.005^n-1) $. [3] (c) The loan is cleared after 60 payments. Find $ M $. [2] (d) Find the total interest paid. [2] Show worked answer

(a) After 1 month: $ 20000(1.005)-M $ [1]. After 2: $ \big(20000(1.005)-M\big)(1.005)-M $, which expands to $ 20000(1.005)^2-M(1.005 + 1) $ [1].
(b) After $ n $ payments: $ 20000(1.005)^n $ $ {}- M(1 + 1.005 + \dots + 1.005^{n-1}) $ [1]. The bracket is a GP sum $ \dfrac{1.005^n-1}{0.005} $ [1], and $ \dfrac{1}{0.005} = 200 $, giving the result [1].
(c) Set the balance to zero with $ n = 60 $: $ M = \dfrac{20000(1.005)^{60}}{200(1.005^{60}-1)} $ [1]. With $ 1.005^{60} \approx 1.348850 $, $ M \approx 386.66 $ dollars [1].
(d) Total repaid $ 60 \times 386.66 = 23199.60 $ [1], so interest $ \approx 3199.60 $ dollars, about $3,200 [1].

Frequently asked questions

How do I tell an arithmetic series from a geometric one?

Subtract and divide neighbouring terms. If the differences are equal, it is arithmetic. If the ratios are equal, it is geometric. In words, “rises by $200” is arithmetic and “rises by 6%” is geometric.

What is the formula for the sum of a geometric series?

$ S_n = \dfrac{a(r^n-1)}{r-1} $, where $ a $ is the first term, $ r $ the common ratio and $ n $ the number of terms. It works for any $ r \ne 1 $. If $ r = 1 $, every term equals $ a $, so the sum is simply $ na $.

When does a geometric series have a sum to infinity?

Only when $ -1 < r < 1 $. Then $ r^n $ shrinks towards zero as $ n $ grows, and the sum settles at $ \dfrac{a}{1-r} $.

Why does my loan answer differ slightly from the bank’s?

Banks usually round each payment to the cent and may adjust the final one, and some charge interest daily rather than monthly. The geometric series model gives the exact payment for the stated monthly rate.

References

Abramson, J. (2021) College Algebra 2e, Sections 9.2–9.4 (arithmetic sequences, geometric sequences, series). Houston, TX: OpenStax. https://openstax.org/details/books/college-algebra-2e
Brealey, R.A., Myers, S.C. and Allen, F. (2020) Principles of Corporate Finance, 13th edn, Chapter 2 (perpetuities and annuities). New York: McGraw-Hill Education.
Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027, Pure Mathematics 1, topic 1.6 Series. Cambridge: Cambridge University Press & Assessment.
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021), SL 1.2, 1.3, 1.4 and 1.8. Cardiff: IBO.
College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description, Unit 10, topic 10.2. New York: College Board.

	Plan A (arithmetic)	Plan B (geometric)
First term	\( a = 3000 \)	\( a = 3000 \)
Rule	\( d = 200 \)	\( r = 1.06 \)
Number of terms	\( n = 10 \) or \( 20 \)	\( n = 10 \) or \( 20 \)
Year \( n \) saving	\( 3000 + 200(n-1) \)	\( 3000(1.06)^{n-1} \)