Properties of Series

Mathematics · FE Reference Handbook section

Mathematics

10 formulas

10 exam-style examples

~60 min

All Mathematics lectures

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 — Properties of Series

A schedule of 31 monthly inspections starts at 12 units and increases by 3 units each month. What is the total over the 31 months?

Given

$a_{1} = 12$
$d = 3$
$n = 31$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 12 + (31 - 1)(3) = 102$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 31(12 + 102)/2 = 1,767$

Answer:

S = 1,767 units

Why the other options are there

372.0 (increment ignored)
3,162 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 2

Arithmetic progression sum — Properties of Series (2)

A schedule of 13 monthly inspections starts at 9 units and increases by 4 units each month. What is the total over the 13 months?

Given

$a_{1} = 9$
$d = 4$
$n = 13$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 9 + (13 - 1)(4) = 57$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 13(9 + 57)/2 = 429.0$

Answer:

S = 429.0 units

Why the other options are there

117.0 (increment ignored)
741.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 3

Arithmetic progression sum — Properties of Series (3)

A schedule of 40 monthly inspections starts at 9 units and increases by 2 units each month. What is the total over the 40 months?

Given

$a_{1} = 9$
$d = 2$
$n = 40$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 9 + (40 - 1)(2) = 87$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 40(9 + 87)/2 = 1,920$

Answer:

S = 1,920 units

Why the other options are there

360.0 (increment ignored)
3,480 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 4

Arithmetic progression sum — Properties of Series (4)

A schedule of 40 monthly inspections starts at 8 units and increases by 9 units each month. What is the total over the 40 months?

Given

$a_{1} = 8$
$d = 9$
$n = 40$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 8 + (40 - 1)(9) = 359$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 40(8 + 359)/2 = 7,340$

Answer:

S = 7,340 units

Why the other options are there

320.0 (increment ignored)
14,360 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 5

Arithmetic progression sum — Properties of Series (5)

A schedule of 34 monthly inspections starts at 4 units and increases by 3 units each month. What is the total over the 34 months?

Given

$a_{1} = 4$
$d = 3$
$n = 34$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 4 + (34 - 1)(3) = 103$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 34(4 + 103)/2 = 1,819$

Answer:

S = 1,819 units

Why the other options are there

136.0 (increment ignored)
3,502 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 6

Arithmetic progression sum — Properties of Series (6)

A schedule of 28 monthly inspections starts at 3 units and increases by 6 units each month. What is the total over the 28 months?

Given

$a_{1} = 3$
$d = 6$
$n = 28$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 3 + (28 - 1)(6) = 165$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 28(3 + 165)/2 = 2,352$

Answer:

S = 2,352 units

Why the other options are there

84 (increment ignored)
4,620 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 7

Arithmetic progression sum — Properties of Series (7)

A schedule of 39 monthly inspections starts at 9 units and increases by 8 units each month. What is the total over the 39 months?

Given

$a_{1} = 9$
$d = 8$
$n = 39$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 9 + (39 - 1)(8) = 313$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 39(9 + 313)/2 = 6,279$

Answer:

S = 6,279 units

Why the other options are there

351.0 (increment ignored)
12,207 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 8

Arithmetic progression sum — Properties of Series (8)

A schedule of 13 monthly inspections starts at 6 units and increases by 5 units each month. What is the total over the 13 months?

Given

$a_{1} = 6$
$d = 5$
$n = 13$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 6 + (13 - 1)(5) = 66$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 13(6 + 66)/2 = 468.0$

Answer:

S = 468.0 units

Why the other options are there

78 (increment ignored)
858.0 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 9

Arithmetic progression sum — Properties of Series (9)

A schedule of 23 monthly inspections starts at 9 units and increases by 9 units each month. What is the total over the 23 months?

Given

$a_{1} = 9$
$d = 9$
$n = 23$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 9 + (23 - 1)(9) = 207$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 23(9 + 207)/2 = 2,484$

Answer:

S = 2,484 units

Why the other options are there

207.0 (increment ignored)
4,761 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series

Example 10

Arithmetic progression sum — Properties of Series (10)

A schedule of 16 monthly inspections starts at 10 units and increases by 7 units each month. What is the total over the 16 months?

Given

$a_{1} = 10$
$d = 7$
$n = 16$

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

Last term
$aₙ = a_{1} + (n - 1)d$
Substituting
$aₙ = 10 + (16 - 1)(7) = 115$
Sum
$S = n(a_{1} + aₙ)/2$
Substituting
$S = 16(10 + 115)/2 = 1,000$

Answer:

S = 1,000 units

Why the other options are there

160.0 (increment ignored)
1,840 (all terms taken at the maximum)

Reference: FE Reference Handbook — Mathematics → Properties of Series