CALCU-(10) Find the n^ th term of the sequence\[2

(10) Find the

n^{\text {th }}

term of the sequence

2,12,36,80,150

(11) (a) Find the gradient of

A B

, whene

A(7,5)

and B>(9,-5) (b) What is the gradient of perpendicular bisector of

A B

.

Solution by Steps

step 1

Identify the pattern in the sequence to find the

n^{\text{th}}

term

step 2

Calculate the differences between the terms:

12 - 2 = 10

,

36 - 12 = 24

,

80 - 36 = 44

,

150 - 80 = 70

step 3

Notice that the differences themselves form a sequence:

10, 24, 44, 70

step 4

Calculate the second set of differences:

24 - 10 = 14

,

44 - 24 = 20

,

70 - 44 = 26

step 5

Observe that the second set of differences is increasing by

6

each time, indicating a quadratic pattern

step 6

The

n^{\text{th}}

term of a quadratic sequence is given by

an^2 + bn + c

. Use the first three terms to form a system of equations to solve for

a

,

b

, and

c

step 7

For the first term (

n=1

), the equation is

a(1)^2 + b(1) + c = 2

step 8

For the second term (

n=2

), the equation is

a(2)^2 + b(2) + c = 12

step 9

For the third term (

n=3

), the equation is

a(3)^2 + b(3) + c = 36

step 10

Solve the system of equations to find

a = 2

,

b = 3

, and

c = -3

step 11

The

n^{\text{th}}

term of the sequence is

2n^2 + 3n - 3

Answer

The

n^{\text{th}}

term of the sequence is

2n^2 + 3n - 3

.

Key Concept

Identifying the pattern in a sequence and solving for the coefficients of a quadratic sequence

Explanation

The sequence is quadratic, and by solving the system of equations formed by the first three terms, we find the coefficients for the quadratic expression representing the

n^{\text{th}}

term.

Solution by Steps

step 1

To find the gradient of

AB

, use the coordinates of points

A(7,5)

and

B(9,-5)

step 2

The gradient (slope) formula is

m = \frac{y_2 - y_1}{x_2 - x_1}

step 3

Substitute the coordinates into the formula:

m = \frac{-5 - 5}{9 - 7}

step 4

Simplify to find the gradient:

m = \frac{-10}{2} = -5

Answer

The gradient of

AB

is

-5

.

Key Concept

Calculating the gradient (slope) of a line given two points

Explanation

The gradient is the vertical change divided by the horizontal change between two points on a line.

Solution by Steps

step 1

The gradient of the perpendicular bisector of

AB

is the negative reciprocal of the gradient of

AB

step 2

Since the gradient of

AB

is

-5

, the gradient of the perpendicular bisector is

\frac{1}{5}

Answer

The gradient of the perpendicular bisector of

AB

is

\frac{1}{5}

.

Key Concept

Finding the gradient of a perpendicular bisector

Explanation

The gradient of the perpendicular bisector is the negative reciprocal of the original line's gradient.

(c) Find the midpoint of

A B

(d) Wite down the equation of perpendicular bisector of

A B

. (12) A regular polygon has an exterior ande of

24^{\circ}

. How many sides are there in the polygon?

Solution by Steps

step 1

To find the midpoint

M

of

AB

, we use the midpoint formula

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

step 2

Given points

A(7,5)

and

B(9,-5)

, we substitute into the midpoint formula:

M = \left( \frac{7 + 9}{2}, \frac{5 + (-5)}{2} \right)

step 3

Simplifying the coordinates of

M

:

M = \left( \frac{16}{2}, \frac{0}{2} \right)

step 4

The midpoint

M

is

M = (8, 0)

Answer

The midpoint

M

of

AB

is

(8, 0)

.

Key Concept

Midpoint of a Line Segment

Explanation

The midpoint of a line segment with endpoints

A(x_1, y_1)

and

B(x_2, y_2)

is found using the midpoint formula

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

.

---

Solution by Steps

step 1

To write down the equation of the perpendicular bisector of

AB

, we first need the slope of

AB

, which is

m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}

step 2

Using points

A(7,5)

and

B(9,-5)

, we calculate

m_{AB} = \frac{-5 - 5}{9 - 7}

step 3

Simplifying, we get

m_{AB} = \frac{-10}{2} = -5

step 4

The slope of the perpendicular bisector is the negative reciprocal of

m_{AB}

, so

m_{perp} = -\frac{1}{m_{AB}} = -\frac{1}{-5} = \frac{1}{5}

step 5

Using the midpoint

M(8, 0)

and slope

m_{perp}

, the equation of the perpendicular bisector is

y - y_1 = m_{perp}(x - x_1)

step 6

Substituting

M(8, 0)

and

m_{perp} = \frac{1}{5}

into the equation, we get

y - 0 = \frac{1}{5}(x - 8)

step 7

Simplifying, we get the equation of the perpendicular bisector:

y = \frac{1}{5}x - \frac{8}{5}

Answer

The equation of the perpendicular bisector of

AB

is

y = \frac{1}{5}x - \frac{8}{5}

.

Key Concept

Perpendicular Bisector of a Line Segment

Explanation

The perpendicular bisector of a line segment is a line that is perpendicular to the segment and passes through its midpoint. The slope of the perpendicular bisector is the negative reciprocal of the original line's slope.

---

Solution by Steps

step 1

The sum of the exterior angles of any polygon is

360^\circ

step 2

If one exterior angle is

24^\circ

, we divide

360^\circ

by

24^\circ

to find the number of sides

step 3

Calculating the number of sides:

\frac{360^\circ}{24^\circ} = 15

Answer

The polygon has 15 sides.

Key Concept

Exterior Angles of a Polygon

Explanation

The sum of the exterior angles of any polygon is always

360^\circ

, regardless of the number of sides. To find the number of sides, divide

360^\circ

by the measure of one exterior angle.

lies between

14.15 \mathrm{~km}

aliu minutes. between 45.5 minutes and 46.5 minutes. of speed

=\frac{\text { upper bound of distance }}{\text { lowerbound of time }}

=\frac{14.25}{45.5} \mathrm{~km} /

minute

=\frac{14.25}{45.5+60} \mathrm{~km} / \mathrm{h}

=18.8 \mathrm{~km} / \mathrm{h}

to 3 significant figures f speed

=\frac{\text { lower bound of distance }}{\text { upper bound of time }}

=\frac{14.15}{46.5+60} \mathrm{~km} / \mathrm{h}

=18.3 \mathrm{~km} / \mathrm{h}

to 3 significant figures Note To find the upper bound of any combined measurement, work out which of the upper and lower bounds of the given measurements you need to use to give you the greatest result. If you aren't sure, experiment! Exercise 10.3 1 Pencils have a width of

8 \mathrm{~mm}

, to the nearest millimetre. What is the smallest total width of 10 pencils? 2 Find the upper and lower bounds of the floor area of a rectangular room with these measurements. a

5.26 \mathrm{~m}

by

3.89 \mathrm{~m}

to the nearest centimetre b

7.6 \mathrm{~m}

by

5.2 \mathrm{~m}

to the nearest

10 \mathrm{~cm}

3 Calculate the upper and lower bounds of the distance travelled for each of these pairs of speed and time. Give your answers to 3 significant figures. a

92.4 \mathrm{~cm} / \mathrm{second}

for 12.3 seconds (both values to 3 s.f.) b

1.54 \mathrm{~m} / \mathrm{second}

for 8.20 seconds (both values to 3 s.f.) c

57 \mathrm{~km} / \mathrm{h}

for 2.5 hours (both values to 2 s.f.) d

5.61 \mathrm{~m} / \mathrm{second}

for 2.08 seconds (both values to 3 s.f.) 4 Find the upper and lower bounds of the average speed for these measured times and distances. Give your answers to 3 significant figures. a

6.4 \mathrm{~cm}

in 1.2 seconds (both values to 2 s.f.) b

106 \mathrm{~m}

in 10.0 seconds (both values to 3 s.f.) Photocopying is prohibited Mr.print, Nawala - 0768486453

Solution by Steps

step 1

To find the smallest total width of 10 pencils, we consider the lower bound of the width of one pencil

step 2

The width of one pencil is given as 8 mm to the nearest millimetre, so the lower bound is

8 - 0.5 = 7.5

mm

step 3

The total width of 10 pencils is

10 \times 7.5

mm

step 4

Calculating the total width:

10 \times 7.5 = 75

mm

Answer

75 mm

Key Concept

Lower Bound Calculation

Explanation

The smallest total width is found by multiplying the lower bound of one pencil's width by the number of pencils.

Solution by Steps

step 1

To find the upper and lower bounds of the floor area of a rectangular room, we calculate the bounds of each dimension

step 2

For 5.26 m by 3.89 m to the nearest centimetre, the lower bound is

5.255

m and

3.885

m, and the upper bound is

5.265

m and

3.895

m

step 3

The lower bound of the area is

5.255 \times 3.885

m²

step 4

The upper bound of the area is

5.265 \times 3.895

m²

step 5

For 7.6 m by 5.2 m to the nearest 10 cm, the lower bound is

7.55

m and

5.15

m, and the upper bound is

7.65

m and

5.25

m

step 6

The lower bound of the area is

7.55 \times 5.15

m²

step 7

The upper bound of the area is

7.65 \times 5.25

m²

Answer

Lower bound for (a):

20.414475

m², Upper bound for (a):

20.513175

m², Lower bound for (b):

38.8825

m², Upper bound for (b):

40.1625

m²

Key Concept

Area Bounds Calculation

Explanation

The bounds of the area are calculated by multiplying the respective bounds of the length and width.

Solution by Steps

step 1

To calculate the upper and lower bounds of the distance travelled, we use the bounds of speed and time

step 2

For 92.4 cm/s for 12.3 seconds, the lower bound of speed is

92.35

cm/s and time is

12.295

s, and the upper bound of speed is

92.45

cm/s and time is

12.305

s

step 3

The lower bound of distance is

92.35 \times 12.295

cm

step 4

The upper bound of distance is

92.45 \times 12.305

cm

step 5

For 1.54 m/s for 8.20 seconds, the lower bound of speed is

1.535

m/s and time is

8.195

s, and the upper bound of speed is

1.545

m/s and time is

8.205

s

step 6

The lower bound of distance is

1.535 \times 8.195

m

step 7

The upper bound of distance is

1.545 \times 8.205

m

step 8

For 57 km/h for 2.5 hours, the lower bound of speed is

56.5

km/h and time is

2.45

h, and the upper bound of speed is

57.5

km/h and time is

2.55

h

step 9

The lower bound of distance is

56.5 \times 2.45

km

step 10

The upper bound of distance is

57.5 \times 2.55

km

step 11

For 5.61 m/s for 2.08 seconds, the lower bound of speed is

5.605

m/s and time is

2.075

s, and the upper bound of speed is

5.615

m/s and time is

2.085

s

step 12

The lower bound of distance is

5.605 \times 2.075

m

step 13

The upper bound of distance is

5.615 \times 2.085

m

Answer

Lower bound for (a):

1135.8925

cm, Upper bound for (a):

1137.6975

cm, Lower bound for (b):

12.573175

m, Upper bound for (b):

12.676225

m, Lower bound for (c):

138.425

km, Upper bound for (c):

146.625

km, Lower bound for (d):

11.65525

m, Upper bound for (d):

11.715075

m

Key Concept

Distance Bounds Calculation

Explanation

The bounds of the distance travelled are calculated by multiplying the respective bounds of speed and time.

Solution by Steps

step 1

To find the upper and lower bounds of the average speed, we divide the bounds of distance by the bounds of time

step 2

For 6.4 cm in 1.2 seconds, the lower bound of distance is

6.35

cm and time is

1.195

s, and the upper bound of distance is

6.45

cm and time is

1.205

s

step 3

The lower bound of average speed is

6.35 / 1.205

cm/s

step 4

The upper bound of average speed is

6.45 / 1.195

cm/s

step 5

For 106 m in 10.0 seconds, the lower bound of distance is

105.5

m and time is

9.995

s, and the upper bound of distance is

106.5

m and time is

10.005

s

step 6

The lower bound of average speed is

105.5 / 10.005

m/s

step 7

The upper bound of average speed is

106.5 / 9.995

m/s

Answer

Lower bound for (a):

5.27029

cm/s, Upper bound for (a):

5.356485

cm/s, Lower bound for (b):

10.55

m/s, Upper bound for (b):

10.65

m/s

Key Concept

Average Speed Bounds Calculation

Explanation

The bounds of the average speed are calculated by dividing the respective bounds of distance by the bounds of time.

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