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CBSE · Class 11 · Mathematics

Infinite Geometric Series (Appendix 1: A.1.3)

3 concepts · 24 practice questions with answers · Free to read

Sum to infinity of a G.P.

From daily life

Fold a ribbon: take half, then half of the rest, then half of that. Though you keep going forever, the total taken never goes beyond the whole ribbon.

Simple definition

For the infinite G.P. a, ar, ar², … with common ratio r, the sum of the first n terms is a(1 − rⁿ)/(1 − r). If |r| < 1, rⁿ becomes almost 0 for large n, so the sum to infinity is S = a/(1 − r). If |r| ≥ 1 there is no finite sum.

Formula: S = a ÷ (1 − r), |r| < 1

Rule 1: Use the formula — Check |r| < 1 first.

  • Easy: 1 + 1/2 + 1/4 + … : a = 1, r = 1/2; S = 2.
  • Medium: 3 + 1 + 1/3 + … : a = 3, r = 1/3; S = 9/2.
  • Exam-type: 6 − 3 + 3/2 − … : a = 6, r = −1/2; S = 4.

Rule 2: Partial sums settle down — Look at the sum of the first few terms.

  • Easy: For a = 1, r = 1/2 the sums are 1, 1.5, 1.75, 1.875, … and approach 2.
  • Medium: For 2 + 4 + 8 + … the sums 2, 6, 14, … keep growing; no sum to infinity.
  • Exam-type: Whether r is positive or negative, the condition is only |r| < 1.
Remember

Check |r| < 1 before using the formula.

Common mistake

Using a/(1 − r) when r = 2 or r = −3.

Memory clue

First over one minus ratio.

Practice questions

Easy
  1. 1Find the sum to infinity of 1 + 1/2 + 1/4 + …
  2. 2Find the sum to infinity of a = 3, r = 1/3.
  3. 3Does 2 + 4 + 8 + … have a sum to infinity?
Medium
  1. 4Find the sum to infinity of 6 − 3 + 3/2 − …
  2. 5Find the sum to infinity of a = 10, r = 0.2.
  3. 6The first term is 5 and the sum to infinity is 15. Find r.
Challenging
  1. 7The sum to infinity is 8 and r = 1/4. Find the first term.
  2. 8Find the sum to infinity of 1/2 + 1/6 + 1/18 + …
Show answers (try them yourself first)
  1. 12
  2. 29/2
  3. 3No, since |r| = 2 is not less than 1
  4. 44
  5. 525/2
  6. 62/3
  7. 76
  8. 83/4

Recurring decimals as fractions

From daily life

The calculator shows 1 ÷ 3 = 0.3333… for ever. Writing it as an infinite G.P. shows why it is exactly 1/3.

Simple definition

A recurring decimal such as 0.4545… = 0.45 + 0.0045 + 0.000045 + … is an infinite G.P. with first term 0.45 and ratio 0.01. Its sum a/(1 − r) is the fraction.

Formula: 0.dddd… = d/10 ÷ (1 − 1/10)

Rule 1: Pure recurring — All digits repeat.

  • Easy: 0.333… : a = 0.3, r = 0.1; S = 1/3.
  • Medium: 0.272727… : a = 0.27, r = 0.01; S = 3/11.
  • Exam-type: 0.4545… : S = 5/11.

Rule 2: Mixed recurring — Split off the non-repeating part.

  • Easy: 0.1222… = 0.1 + 0.0222… = 1/10 + 2/90 = 11/90.
  • Medium: 2.7272… = 2 + 8/11 = 30/11.
  • Exam-type: 0.999… : a = 0.9, r = 0.1; S = 1, that is exactly 1.
Remember

Block of k repeating digits: ratio is 1/10^k.

Common mistake

Taking the first term as the whole repeating block times 10 instead of the block itself.

Memory clue

Repeating block over nines.

Practice questions

Easy
  1. 1Write 0.555… as a fraction.
  2. 2Write 0.0909… as a fraction.
  3. 3Write 0.4545… as a fraction.
Medium
  1. 4Write 0.1222… as a fraction.
  2. 5Write 0.3636… as a fraction.
  3. 6Write 0.8333… as a fraction.
Challenging
  1. 7Write 2.7272… as a fraction.
  2. 8Use an infinite G.P. to show that 0.999… = 1.
Show answers (try them yourself first)
  1. 15/9
  2. 21/11
  3. 35/11
  4. 411/90
  5. 54/11
  6. 65/6
  7. 730/11
  8. 8a = 0.9, r = 0.1; S = 0.9/0.9 = 1

Infinite G.P. in everyday problems

From daily life

A ball dropped on the floor bounces again and again, each bounce lower than the last. It looks endless, yet the total distance is a fixed number.

Simple definition

Whenever a quantity repeats in a fixed fraction of the previous one, the total can be found with S = a/(1 − r). For a ball dropped from height h that rebounds to the fraction r of its previous height, the total distance travelled is h(1 + r)/(1 − r).

Formula: Total distance = h(1 + r) ÷ (1 − r)

Rule 1: Bouncing ball — Down once, then up and down for each rebound.

  • Easy: h = 10 m, r = 0.6: total = 10 × 1.6 ÷ 0.4 = 40 m.
  • Medium: h = 8 m, r = 1/2: total = 8 × 1.5 ÷ 0.5 = 24 m.
  • Exam-type: Only the upward distances, h = 10, r = 0.6: 6 + 3.6 + … = 6 ÷ 0.4 = 15 m.

Rule 2: Shrinking shapes and doses — Find a and r, then S.

  • Easy: A swing of 20 cm, each next swing 0.9 of the last: total path = 20 ÷ 0.1 = 200 cm.
  • Medium: 100 mg dose daily, 1/4 of it remains after a day: long-run amount just after a dose = 100 ÷ 0.75 ≈ 133.33 mg.
  • Exam-type: Squares of side 8 cm, each new square joins the midpoints of the previous: areas 64, 32, 16, … add up to 128 cm².
Remember

Count the first drop separately, then the up-and-down pairs.

Common mistake

Forgetting that each rebound is travelled twice (up and down).

Memory clue

First drop plus twice the rebounds.

Practice questions

Easy
  1. 1A ball dropped from 10 m rebounds to 0.6 of its previous height each time. Find the total distance travelled.
  2. 2A ball dropped from 8 m rebounds to 1/2 of the previous height. Find the total distance.
  3. 3A ball dropped from 20 m rebounds to 0.8 of its previous height. Find the total distance.
Medium
  1. 4A ball dropped from 10 m rebounds to 0.6 of its previous height. Find the total of the upward distances only.
  2. 5The first swing of a pendulum is 20 cm and each swing is 0.9 of the one before. Find the total path.
  3. 6A patient takes 100 mg daily and 1/4 of the drug remains in the body after one day. Find the amount in the body just after a dose in the long run.
Challenging
  1. 7Squares are drawn by joining midpoints of sides, starting with a square of side 8 cm. Find the sum of all the areas.
  2. 8A ball dropped from h metres rebounds to 3/5 of its previous height and travels 80 m in total. Find h.
Show answers (try them yourself first)
  1. 140 m
  2. 224 m
  3. 3180 m
  4. 415 m
  5. 5200 cm
  6. 6about 133.33 mg
  7. 7128 cm²
  8. 820 m