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Quantitative Aptitude · Sequence and Series

Sum of Infinite GP and Geometric Mean for CA Foundation

Updated 1 October 2026

An infinite GP has a finite sum only when its common ratio r satisfies |r| < 1. Then the sum to infinity is S∞ = a ÷ (1 − r). The geometric mean of two positive numbers a and b is √(ab). To insert n GMs between them, find the common ratio r = (b ÷ a)^(1/(n+1)).

Understand Sum of Infinite GP and Geometric Mean

A geometric progression (GP) is a sequence where each term is the previous term multiplied by a fixed number r, the common ratio. If the GP never stops, it is an infinite GP: a, ar, ar², ar³, ...

Will the sum of infinitely many terms be a finite number? It depends on r. If |r| ≥ 1, the terms do not shrink, so the sum keeps growing or swings without settling. If |r| < 1, each term is smaller than the last in size, and the terms fade toward zero. The total then settles to a fixed value. This is called convergence.

For |r| < 1, the sum of the first n terms is a(1 − rⁿ) ÷ (1 − r). As n becomes very large, rⁿ becomes almost 0. So the sum approaches a ÷ (1 − r). This is why the formula needs |r| < 1, that is, −1 < r < 1.

The geometric mean of two positive numbers a and b is the middle term of a GP a, G, b. Since G ÷ a = b ÷ G, we get G² = ab and G = √(ab). To insert n geometric means between a and b, you build a GP with n + 2 terms whose first term is a and last term is b. Then find r and write the terms.

A recurring decimal is an infinite GP in disguise. For example, 0.272727... = 0.27 + 0.0027 + 0.000027 + ..., with first term 0.27 and ratio 0.01. Use the sum formula to get a fraction.

Key formulas to remember

Condition for convergence
|r| < 1, i.e. −1 < r < 1
The sum to infinity exists only under this condition. Check it before using the formula.
Sum to infinity of a GP
S∞ = a ÷ (1 − r)
a is the first term and r the common ratio, with |r| < 1.
Sum of first n terms
Sₙ = a(1 − rⁿ) ÷ (1 − r), for r ≠ 1
The finite sum. As n grows with |r| < 1, rⁿ tends to 0.
Geometric mean of two numbers
G = √(ab)
Use for positive a and b. If both are negative, G is taken as −√(ab) so that a, G, b is a GP.
Common ratio when inserting n GMs
r = (b ÷ a)^(1/(n+1))
The GP has n + 2 terms, so b = a·r^(n+1).
Product of n GMs inserted between a and b
G₁ × G₂ × ... × Gₙ = (√(ab))ⁿ
Valid for positive a and b. Terms equally far from the two ends multiply to ab.
Pure recurring decimal
Pure recurring decimal 0.(block) = block ÷ (10^k − 1), k = block length
Example: 0.272727... = 27 ÷ 99 = 3/11.

How to solve Sum of Infinite GP and Geometric Mean questions

Use this method for any question on infinite GP sums, recurring decimals or geometric means.

  1. 1Identify the first term a and the common ratio r = second term ÷ first term.
  2. 2Check the condition |r| < 1. If it fails, the infinite sum does not exist.
  3. 3Substitute into S∞ = a ÷ (1 − r). Handle fractions carefully and simplify.
  4. 4For a recurring decimal, split it into a non-repeating part and a repeating part. Treat the repeating part as a GP with ratio 1/10^k.
  5. 5For one GM between a and b, compute G = √(ab). Check that the numbers are positive.
  6. 6For n GMs, find r = (b ÷ a)^(1/(n+1)), then the terms are ar, ar², ..., arⁿ.
  7. 7If a question gives S∞ and one other fact, form an equation in a and r and solve.
  8. 8Match the result with the options and check it is sensible. For example, for a > 0 and 0 < r < 1, the sum is larger than a.

Quickest way: Shortcut using S∞ = first term ÷ (1 − ratio)

When to use it: Use for any MCQ asking the sum of an infinite series, a recurring decimal or a GM, especially when options differ clearly.

  1. Write only a and r. Compute r by dividing term 2 by term 1.
  2. If |r| ≥ 1, mark 'does not exist' or 'infinite' if that option is given.
  3. Compute a ÷ (1 − r) directly. For r = 1/k, S∞ = a·k ÷ (k − 1).
  4. For a negative r, remember 1 − r is larger than 1, so for a > 0 the sum is smaller than a.
  5. For a pure recurring decimal, write the repeating block over 9s: one digit over 9, two digits over 99.
  6. For a GM of two numbers, test options by squaring: the correct G has G² = ab.
  7. The GM of a and 4a (a > 0) is 2a. Spot such patterns fast.
  8. If stuck on a long question, skip it. A wrong answer costs 0.25 marks.

Common mistakes in Sum of Infinite GP and Geometric Mean

  • Using S∞ = a ÷ (1 − r) when |r| ≥ 1

    Students memorise the formula and forget the condition.

    Fix: Always compute r first and check −1 < r < 1 before applying the formula.

  • Taking r as the wrong ratio, such as the difference of terms

    Mixing up AP and GP habits.

    Fix: In a GP, r = term 2 ÷ term 1. Confirm with term 3 ÷ term 2.

  • Forgetting the sign of r in an alternating series

    Students see 1, −1/2, 1/4 and use r = 1/2.

    Fix: Here r = −1/2, so S∞ = 1 ÷ (1 + 1/2) = 2/3. Keep the sign.

  • Writing the GM as (a + b) ÷ 2

    Confusing geometric mean with arithmetic mean.

    Fix: GM is √(ab). AM is (a + b) ÷ 2. For positive unequal numbers, GM is smaller than AM.

  • Using n instead of n + 1 as the root when inserting n means

    Forgetting that n means plus two end terms give n + 1 steps.

    Fix: Use r^(n+1) = b ÷ a. Inserting 3 GMs means taking the 4th root.

  • Forgetting the non-repeating part in a mixed recurring decimal

    Treating 0.1666... as if every digit repeats.

    Fix: Split it: 0.1 + 0.0666... Apply the GP formula only to the repeating part, then add.

Worked examples

Example 1

The sum to infinity of the series 8 + 4 + 2 + 1 + ... is: (A) 12 (B) 14 (C) 16 (D) 32

Show the solution
  1. First term a = 8. Common ratio r = 4 ÷ 8 = 1/2.
  2. |r| = 1/2 < 1, so the sum exists.
  3. S∞ = a ÷ (1 − r) = 8 ÷ (1 − 1/2) = 8 ÷ (1/2) = 16.

Answer: (C) 16

Example 2

The value of the recurring decimal 0.545454... as a fraction in lowest terms is: (A) 6/11 (B) 5/9 (C) 27/50 (D) 54/101

Show the solution
  1. Write 0.545454... = 0.54 + 0.0054 + 0.000054 + ...
  2. First term a = 0.54, ratio r = 0.0054 ÷ 0.54 = 0.01. Since |r| < 1, the sum exists.
  3. S∞ = 0.54 ÷ (1 − 0.01) = 0.54 ÷ 0.99 = 54/99.
  4. Divide numerator and denominator by 9: 54/99 = 6/11.

Answer: (A) 6/11

Example 3

Three geometric means are inserted between 3 and 48. The second geometric mean is: (A) 6 (B) 12 (C) 24 (D) 18

Show the solution
  1. Here a = 3, b = 48, n = 3. The GP has 5 terms, so b = a·r⁴.
  2. r⁴ = 48 ÷ 3 = 16, so r = 2 (taking the positive real root).
  3. The terms are 3, 6, 12, 24, 48.
  4. The three GMs are 6, 12, 24. The second is 12.
  5. Check: the middle term of the 5-term GP is √(3 × 48) = √144 = 12.

Answer: (B) 12

Exam tips

  • Questions often hide the series in words or a sigma form. Write the first two terms to find a and r.
  • Recurring decimal questions are quick marks. Use block over 9s for pure recurring decimals.
  • When options include 'does not exist', check |r| < 1 first.
  • For inserting means, count the terms carefully: n means give n + 2 terms and n + 1 ratio steps.
  • Use the middle-term check G² = ab to confirm a GM in seconds.

Practice questions from Sequence and Series

Sum of Infinite GP and Geometric Mean: frequently asked questions

When does the sum of an infinite GP exist?

It exists only when the common ratio r lies between −1 and 1, that is |r| < 1. In that case the terms shrink toward zero and the sum settles to a ÷ (1 − r). For |r| ≥ 1 the sum does not exist as a finite number.

What is the formula for the geometric mean between two numbers?

For positive numbers a and b, the geometric mean is G = √(ab). It is the middle term of the GP a, G, b. It is always less than or equal to the arithmetic mean for positive numbers.

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

Write the decimal as a sum of terms with a repeating block. Take the first block as a and 1/10^k as r, where k is the block length. Apply S∞ = a ÷ (1 − r) and simplify. For a mixed decimal, add the non-repeating part separately.

How many geometric means can be inserted between two numbers?

Any number n of them, as long as a real GP exists. You find r = (b ÷ a)^(1/(n+1)). Then the means are ar, ar², ..., arⁿ.