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Geometric Progression

Mathematics · FE Reference Handbook section

Mathematics
4 formulas
10 exam-style examples
~53 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • To determine whether a given finite sequence is a geometric progression (G.P.), divide each number after the first by the
  • preceding number. If the quotients are equal, the series is geometric:
  • 4. The last or nth term is l.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Arithmetic progression sum — Geometric Progression

A schedule of 12 monthly inspections starts at 3 units and increases by 3 units each month. What is the total over the 12 months?

Given

  • a1=3a_{1} = 3
  • d=3d = 3
  • n=12n = 12

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=3+(12−1)(3)=36aₙ = 3 + (12 - 1)(3) = 36

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=12(3+36)/2=234.0S = 12(3 + 36)/2 = 234.0
Answer:
S=234.0unitsS = 234.0 units

Why the other options are there

  • 36 (increment ignored)
  • 432.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 2
Arithmetic progression sum — Geometric Progression (2)

A schedule of 40 monthly inspections starts at 6 units and increases by 6 units each month. What is the total over the 40 months?

Given

  • a1=6a_{1} = 6
  • d=6d = 6
  • n=40n = 40

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=6+(40−1)(6)=240aₙ = 6 + (40 - 1)(6) = 240

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=40(6+240)/2=4,920S = 40(6 + 240)/2 = 4,920
Answer:
S=4,920unitsS = 4,920 units

Why the other options are there

  • 240.0 (increment ignored)
  • 9,600 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 3
Arithmetic progression sum — Geometric Progression (3)

A schedule of 15 monthly inspections starts at 3 units and increases by 9 units each month. What is the total over the 15 months?

Given

  • a1=3a_{1} = 3
  • d=9d = 9
  • n=15n = 15

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=3+(15−1)(9)=129aₙ = 3 + (15 - 1)(9) = 129

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=15(3+129)/2=990.0S = 15(3 + 129)/2 = 990.0
Answer:
S=990.0unitsS = 990.0 units

Why the other options are there

  • 45 (increment ignored)
  • 1,935 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 4
Arithmetic progression sum — Geometric Progression (4)

A schedule of 38 monthly inspections starts at 5 units and increases by 5 units each month. What is the total over the 38 months?

Given

  • a1=5a_{1} = 5
  • d=5d = 5
  • n=38n = 38

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=5+(38−1)(5)=190aₙ = 5 + (38 - 1)(5) = 190

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=38(5+190)/2=3,705S = 38(5 + 190)/2 = 3,705
Answer:
S=3,705unitsS = 3,705 units

Why the other options are there

  • 190.0 (increment ignored)
  • 7,220 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 5
Arithmetic progression sum — Geometric Progression (5)

A schedule of 25 monthly inspections starts at 9 units and increases by 7 units each month. What is the total over the 25 months?

Given

  • a1=9a_{1} = 9
  • d=7d = 7
  • n=25n = 25

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=9+(25−1)(7)=177aₙ = 9 + (25 - 1)(7) = 177

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=25(9+177)/2=2,325S = 25(9 + 177)/2 = 2,325
Answer:
S=2,325unitsS = 2,325 units

Why the other options are there

  • 225.0 (increment ignored)
  • 4,425 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 6
Arithmetic progression sum — Geometric Progression (6)

A schedule of 20 monthly inspections starts at 5 units and increases by 5 units each month. What is the total over the 20 months?

Given

  • a1=5a_{1} = 5
  • d=5d = 5
  • n=20n = 20

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=5+(20−1)(5)=100aₙ = 5 + (20 - 1)(5) = 100

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=20(5+100)/2=1,050S = 20(5 + 100)/2 = 1,050
Answer:
S=1,050unitsS = 1,050 units

Why the other options are there

  • 100.0 (increment ignored)
  • 2,000 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 7
Arithmetic progression sum — Geometric Progression (7)

A schedule of 14 monthly inspections starts at 12 units and increases by 6 units each month. What is the total over the 14 months?

Given

  • a1=12a_{1} = 12
  • d=6d = 6
  • n=14n = 14

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=12+(14−1)(6)=90aₙ = 12 + (14 - 1)(6) = 90

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=14(12+90)/2=714.0S = 14(12 + 90)/2 = 714.0
Answer:
S=714.0unitsS = 714.0 units

Why the other options are there

  • 168.0 (increment ignored)
  • 1,260 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 8
Arithmetic progression sum — Geometric Progression (8)

A schedule of 22 monthly inspections starts at 5 units and increases by 5 units each month. What is the total over the 22 months?

Given

  • a1=5a_{1} = 5
  • d=5d = 5
  • n=22n = 22

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=5+(22−1)(5)=110aₙ = 5 + (22 - 1)(5) = 110

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=22(5+110)/2=1,265S = 22(5 + 110)/2 = 1,265
Answer:
S=1,265unitsS = 1,265 units

Why the other options are there

  • 110.0 (increment ignored)
  • 2,420 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 9
Arithmetic progression sum — Geometric Progression (9)

A schedule of 20 monthly inspections starts at 7 units and increases by 2 units each month. What is the total over the 20 months?

Given

  • a1=7a_{1} = 7
  • d=2d = 2
  • n=20n = 20

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=7+(20−1)(2)=45aₙ = 7 + (20 - 1)(2) = 45

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=20(7+45)/2=520.0S = 20(7 + 45)/2 = 520.0
Answer:
S=520.0unitsS = 520.0 units

Why the other options are there

  • 140.0 (increment ignored)
  • 900.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

Example 10
Arithmetic progression sum — Geometric Progression (10)

A schedule of 30 monthly inspections starts at 9 units and increases by 9 units each month. What is the total over the 30 months?

Given

  • a1=9a_{1} = 9
  • d=9d = 9
  • n=30n = 30

Find

Total (sum of the series)

Start with the thinking

  • Constant increment means an arithmetic series.
  • Get the last term first, then use the pair-average form.

Step-by-step solution

  1. Last term

    an=a1+(n−1)daₙ = a_{1} + (n - 1)d

  2. Substituting

    an=9+(30−1)(9)=270aₙ = 9 + (30 - 1)(9) = 270

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=30(9+270)/2=4,185S = 30(9 + 270)/2 = 4,185
Answer:
S=4,185unitsS = 4,185 units

Why the other options are there

  • 270.0 (increment ignored)
  • 8,100 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Geometric Progression

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