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Fundamentals of Business Mathematics and Statistics · Arithmetic Progression and Geometric Progression

Sum of GP and Sum to Infinity: Formulas and Method

Updated 10 October 2026 · Fact-checked

The sum of n terms of a GP with first term a and common ratio r is a(rⁿ − 1) ÷ (r − 1) when r ≠ 1, and na when r = 1. If |r| < 1, the infinite GP adds up to a ÷ (1 − r). Find a, r and n, check the condition, then substitute.

Understand Sum of GP and Sum to Infinity

A geometric progression (GP) is a list of numbers where each term is the previous term multiplied by a fixed number r, called the common ratio. For 3, 6, 12, 24, the first term a is 3 and r is 2.

To add the first n terms, you do not need to add one by one. Write the sum Sn = a + ar + ar² + ... + arⁿ⁻¹. Multiply it by r and subtract. Almost everything cancels, leaving Sn(1 − r) = a(1 − rⁿ). That is where the formula comes from.

Now think about adding terms forever. If r is bigger than 1 in size, the terms keep growing and the sum has no limit. If |r| < 1 (r lies between −1 and 1), each term is smaller than the one before. The term rⁿ shrinks towards zero as n grows, so the sum settles at a fixed value. Putting rⁿ = 0 in the formula gives the sum to infinity: S∞ = a ÷ (1 − r).

This is also how a recurring decimal becomes a fraction. A decimal such as 0.272727... is 0.27 + 0.0027 + 0.000027 + ..., which is an infinite GP with a = 0.27 and r = 0.01. Its sum is the fraction you want.

The infinite sum exists only when |r| < 1. A negative r with |r| < 1 is allowed, and the sum still works.

Key formulas to remember

nth term of a GP
Tn = a × rⁿ⁻¹
Use it to find n or the last term. The power is n − 1, not n.
Sum of n terms (form for |r| > 1)
Sn = a(rⁿ − 1) ÷ (r − 1)
Valid for any r ≠ 1. Prefer this form when |r| > 1, for example r = 2 or r = −2, as it keeps the working simple.
Sum of n terms (form for |r| < 1)
Sn = a(1 − rⁿ) ÷ (1 − r)
Valid for any r ≠ 1 and gives the same value as the other form. Prefer it when |r| < 1, for example when r is a fraction, as it avoids negative signs.
Sum when r = 1
Sn = n × a
Every term equals a. The formulas above cannot be used because the denominator becomes zero.
Sum to infinity
S∞ = a ÷ (1 − r), valid only when |r| < 1
If |r| ≥ 1, the infinite sum does not exist as a fixed number.
Common ratio
r = T2 ÷ T1 = T3 ÷ T2
Divide any term by the one before it. It can be negative or a fraction.

How to solve Sum of GP and Sum to Infinity questions

Use this order for any question on the sum of a GP, finite or infinite.

  1. 1Write down the first term a and find r by dividing the second term by the first. Check with the third term divided by the second.
  2. 2Decide the type: a fixed number of terms (use Sn), or 'to infinity' or a recurring decimal (use S∞).
  3. 3For S∞, check |r| < 1 first. If it fails, the sum does not exist.
  4. 4For Sn, if r = 1 the answer is na. Otherwise both forms are valid for any r ≠ 1. Use a(rⁿ − 1) ÷ (r − 1) when |r| > 1, and a(1 − rⁿ) ÷ (1 − r) when |r| < 1.
  5. 5Substitute carefully. Work out rⁿ separately and keep brackets around negative values of r.
  6. 6Simplify the fraction. If the question asks for n, set Sn equal to the given value and solve for rⁿ.
  7. 7Match your answer with the options and check that it is sensible. For r > 1 the sum is larger than the last term. For |r| < 1 the infinite sum is close to a.

Quickest way: Spot a, r, then plug in

When to use it: Use this for most MCQs where the series is given in a line or a decimal is recurring.

  1. Find r in one line: second term ÷ first term.
  2. If the series has few terms (up to 4 or 5), adding them directly can be faster than the formula, for example 2 + 6 + 18 + 54 = 80.
  3. For infinite sums, remember S∞ = first term ÷ (1 − r). For 8 + 4 + 2 + ..., r = ½ and the sum is 8 ÷ ½ = 16.
  4. For a pure recurring decimal like 0.ababab..., the answer is simply ab ÷ 99, and for 0.abc recurring it is abc ÷ 999.
  5. For a decimal with a non-repeating part, split it. Example: 0.1666... = 0.1 + 0.0666... = 1/10 + 0.06 ÷ 0.9 = 1/10 + 1/15 = 1/6.
  6. Test the options: if the infinite sum has a = 6 and r = ⅓, the answer must be bigger than 6, so you can drop smaller options at once.

Common mistakes in Sum of GP and Sum to Infinity

  • Using a ÷ (1 − r) when |r| ≥ 1.

    Students memorise the formula without the condition, so they put in r = 2 and get a negative 'sum'.

    Fix: Always check |r| < 1 first. If it fails, the infinite sum does not exist.

  • Taking the wrong number of terms, using rⁿ⁻¹ instead of rⁿ in the sum formula.

    The nth term uses n − 1 as the power, and this gets mixed up with the sum formula.

    Fix: Remember: nth term has rⁿ⁻¹, but sum of n terms has rⁿ. Test with n = 1: the sum should equal a.

  • Using the formula when r = 1 and dividing by zero.

    A series like 5, 5, 5, ... is not recognised as a GP with r = 1.

    Fix: If r = 1, write Sn = na straight away.

  • Dropping the sign when r is negative.

    For 1 − ½ + ¼ − ..., students write r = ½ instead of −½, and then 1 − r becomes ½ rather than 3/2.

    Fix: Divide the second term by the first with its sign. Here r = −½, so S∞ = 1 ÷ (1 + ½) = 2/3.

  • Wrong first term and ratio for a recurring decimal.

    Students take a = 0.2 for 0.2727... or take r = 0.1 instead of 0.01 for a two-digit repeating block.

    Fix: Take one full repeating block as a. The ratio is 1 divided by 10 raised to the length of the block: 0.01 for two digits, 0.001 for three digits.

  • Making errors when solving for n.

    Students try to guess n or mix the sum with the nth term.

    Fix: Put Sn in the formula, solve for rⁿ, and write that number as a power of r. Check by adding the terms.

Worked examples

Example 1

Find the sum of the first 6 terms of the GP 3, 6, 12, 24, ...

Show the solution
  1. First term a = 3. Common ratio r = 6 ÷ 3 = 2. Number of terms n = 6.
  2. Since |r| > 1, use Sn = a(rⁿ − 1) ÷ (r − 1).
  3. r⁶ = 2⁶ = 64.
  4. S6 = 3 × (64 − 1) ÷ (2 − 1) = 3 × 63 ÷ 1 = 189.
  5. Check by adding: 3 + 6 + 12 + 24 + 48 + 96 = 189.

Answer: 189

Example 2

Express the recurring decimal 0.272727... as a fraction using the sum of an infinite GP.

Show the solution
  1. Write 0.272727... = 0.27 + 0.0027 + 0.000027 + ...
  2. This is a GP with first term a = 0.27 and common ratio r = 0.0027 ÷ 0.27 = 0.01.
  3. Since |r| = 0.01 < 1, the infinite sum exists: S∞ = a ÷ (1 − r).
  4. S∞ = 0.27 ÷ (1 − 0.01) = 0.27 ÷ 0.99 = 27/99.
  5. Divide top and bottom by 9: 27/99 = 3/11.

Answer: 3/11

Exam tips

  • Look at the condition word 'to infinity' or a recurring decimal. That tells you immediately to use a ÷ (1 − r).
  • For a recurring decimal with a pure repeating block, ab ÷ 99 style shortcuts save time, but know the GP method so you can handle mixed decimals.
  • If options look close, find r and check its sign first. Sign errors are the usual reason for wrong options.
  • Questions asking for n are common. Work out rⁿ and then write it as a power, such as 256 = 2⁸.
  • There is no negative marking, so never leave a GP question blank. Eliminate options that break basic logic, such as an infinite sum smaller than a when r is positive.

Practice questions from Arithmetic Progression and Geometric Progression

Sum of GP and Sum to Infinity: frequently asked questions

What is the formula for the sum of n terms of a GP?

For first term a and common ratio r, Sn = a(rⁿ − 1) ÷ (r − 1) when r ≠ 1. You can also write it as a(1 − rⁿ) ÷ (1 − r). Both give the same value. If r = 1, Sn = na.

When does the sum of an infinite GP exist?

It exists only when the common ratio is less than 1 in size, that is, −1 < r < 1. Then S∞ = a ÷ (1 − r). If |r| is 1 or more, the terms do not shrink, so there is no fixed sum.

How do I convert a recurring decimal to a fraction using GP?

Write the decimal as a sum of one repeating block, then the same block shifted, and so on. Take the first block as a and 1 ÷ 10 raised to the block length as r. Then use a ÷ (1 − r) and simplify. For 0.27 recurring, a = 0.27, r = 0.01 and the answer is 3/11.

Can the common ratio be negative in a GP sum?

Yes. The formulas work for negative r. For the infinite sum you still need |r| < 1. For example, 1 − ½ + ¼ − ... has r = −½ and sums to 2/3.