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Arithmetic Progression

Mathematics · FE Reference Handbook section

Mathematics
2 formulas
10 exam-style examples
~49 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 of numbers is an arithmetic progression, subtract each number from the following
  • number. If the differences are equal, the series is arithmetic.
  • 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 — Arithmetic Progression

A schedule of 32 monthly inspections starts at 6 units and increases by 5 units each month. What is the total over the 32 months?

Given

  • a1=6a_{1} = 6
  • d=5d = 5
  • n=32n = 32

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+(32−1)(5)=161aₙ = 6 + (32 - 1)(5) = 161

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=32(6+161)/2=2,672S = 32(6 + 161)/2 = 2,672
Answer:
S=2,672unitsS = 2,672 units

Why the other options are there

  • 192.0 (increment ignored)
  • 5,152 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 2
Arithmetic progression sum — Arithmetic Progression (2)

A schedule of 19 monthly inspections starts at 7 units and increases by 5 units each month. What is the total over the 19 months?

Given

  • a1=7a_{1} = 7
  • d=5d = 5
  • n=19n = 19

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+(19−1)(5)=97aₙ = 7 + (19 - 1)(5) = 97

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=19(7+97)/2=988.0S = 19(7 + 97)/2 = 988.0
Answer:
S=988.0unitsS = 988.0 units

Why the other options are there

  • 133.0 (increment ignored)
  • 1,843 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 3
Arithmetic progression sum — Arithmetic Progression (3)

A schedule of 14 monthly inspections starts at 5 units and increases by 2 units each month. What is the total over the 14 months?

Given

  • a1=5a_{1} = 5
  • d=2d = 2
  • 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=5+(14−1)(2)=31aₙ = 5 + (14 - 1)(2) = 31

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=14(5+31)/2=252.0S = 14(5 + 31)/2 = 252.0
Answer:
S=252.0unitsS = 252.0 units

Why the other options are there

  • 70 (increment ignored)
  • 434.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 4
Arithmetic progression sum — Arithmetic Progression (4)

A schedule of 30 monthly inspections starts at 7 units and increases by 5 units each month. What is the total over the 30 months?

Given

  • a1=7a_{1} = 7
  • d=5d = 5
  • 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=7+(30−1)(5)=152aₙ = 7 + (30 - 1)(5) = 152

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=30(7+152)/2=2,385S = 30(7 + 152)/2 = 2,385
Answer:
S=2,385unitsS = 2,385 units

Why the other options are there

  • 210.0 (increment ignored)
  • 4,560 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 5
Arithmetic progression sum — Arithmetic Progression (5)

A schedule of 23 monthly inspections starts at 6 units and increases by 8 units each month. What is the total over the 23 months?

Given

  • a1=6a_{1} = 6
  • d=8d = 8
  • n=23n = 23

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+(23−1)(8)=182aₙ = 6 + (23 - 1)(8) = 182

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=23(6+182)/2=2,162S = 23(6 + 182)/2 = 2,162
Answer:
S=2,162unitsS = 2,162 units

Why the other options are there

  • 138.0 (increment ignored)
  • 4,186 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 6
Arithmetic progression sum — Arithmetic Progression (6)

A schedule of 37 monthly inspections starts at 9 units and increases by 8 units each month. What is the total over the 37 months?

Given

  • a1=9a_{1} = 9
  • d=8d = 8
  • n=37n = 37

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+(37−1)(8)=297aₙ = 9 + (37 - 1)(8) = 297

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=37(9+297)/2=5,661S = 37(9 + 297)/2 = 5,661
Answer:
S=5,661unitsS = 5,661 units

Why the other options are there

  • 333.0 (increment ignored)
  • 10,989 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 7
Arithmetic progression sum — Arithmetic Progression (7)

A schedule of 38 monthly inspections starts at 9 units and increases by 8 units each month. What is the total over the 38 months?

Given

  • a1=9a_{1} = 9
  • d=8d = 8
  • 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=9+(38−1)(8)=305aₙ = 9 + (38 - 1)(8) = 305

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=38(9+305)/2=5,966S = 38(9 + 305)/2 = 5,966
Answer:
S=5,966unitsS = 5,966 units

Why the other options are there

  • 342.0 (increment ignored)
  • 11,590 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 8
Arithmetic progression sum — Arithmetic Progression (8)

A schedule of 37 monthly inspections starts at 6 units and increases by 5 units each month. What is the total over the 37 months?

Given

  • a1=6a_{1} = 6
  • d=5d = 5
  • n=37n = 37

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+(37−1)(5)=186aₙ = 6 + (37 - 1)(5) = 186

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=37(6+186)/2=3,552S = 37(6 + 186)/2 = 3,552
Answer:
S=3,552unitsS = 3,552 units

Why the other options are there

  • 222.0 (increment ignored)
  • 6,882 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 9
Arithmetic progression sum — Arithmetic Progression (9)

A schedule of 25 monthly inspections starts at 11 units and increases by 3 units each month. What is the total over the 25 months?

Given

  • a1=11a_{1} = 11
  • d=3d = 3
  • 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=11+(25−1)(3)=83aₙ = 11 + (25 - 1)(3) = 83

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=25(11+83)/2=1,175S = 25(11 + 83)/2 = 1,175
Answer:
S=1,175unitsS = 1,175 units

Why the other options are there

  • 275.0 (increment ignored)
  • 2,075 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

Example 10
Arithmetic progression sum — Arithmetic Progression (10)

A schedule of 30 monthly inspections starts at 10 units and increases by 6 units each month. What is the total over the 30 months?

Given

  • a1=10a_{1} = 10
  • d=6d = 6
  • 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=10+(30−1)(6)=184aₙ = 10 + (30 - 1)(6) = 184

  3. Sum

    S=n(a1+an)/2S = n(a_{1} + aₙ)/2

  4. Substituting

    S=30(10+184)/2=2,910S = 30(10 + 184)/2 = 2,910
Answer:
S=2,910unitsS = 2,910 units

Why the other options are there

  • 300.0 (increment ignored)
  • 5,520 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Arithmetic Progression

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