Reduce Non-Linear Function To Linear Function – Examples (A) To (F)

Examples:
Reduce each of the following equations to the linear form. Hence, state the gradient and the Y-intercept of the linear equations in terms of  a and b.
(a)  y = a x 3 + b x 2
(b)  y = a x + b x
(c)  y = a x b x 2
(d)  x y = p x + q x
(e)  y = a x + b x
(f)  a y = b x + 1

[Note :
X and Y cannot have constant, but can have the variables (for example x and y)
 m and c can only have the constant(for example a and b), cannot have the variables x and y]

Solution:







The Line Of Best Fit

2.2 The Line of Best Fit

Line of best fit 
has 2 characteristics:
(i) it passes through as many points as possible,
(ii) the number of points which are not on the line of best fit are equally distributed on the both sides of the line.

Example
Check whether the following graph has the characteristic of the line of best fit

(a)


Yes! This is the Line of Best Fit
.
Notice that you have 3 points passes through the straight line,1 points above the line and 1 point below the line. The number of points which are not on the line of best fit are equally distributed on the both sides of the line.

(b)


No. This is NOT the Line of Best Fit
.
Notice you have more points above the line than below the line. The number of points which are not on the line of best fit are NOT equally distributed on the both sides of the

SPM Practice 3 (Linear Law) – Question 2


Question 2 (10 marks):
Use a graph to answer this question.
Table shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of y 2 x  against  1 x is plotted.


(a) Based on Table, construct a table for the values of 1 x  and  y 2 x .  
( b ) Plot  y 2 x  against  1 x , using a scale of 2 cm to 0.1 unit on the  1 x -axis   and 2cm to 2 units on the  y 2 x -axis.   Hence, draw the line of best fit.

(c) Using the graph in 11(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.

Solution:
(a)


(b)



(c)(i)
When x=2.7,  1 x =0.37 From graph, y 2 x =5.2 y 2 2.7 =5.2 y=3.75


(c)(ii)

Form graph, y-intercept, c = –4 gradient, m= 16( 4 ) 0.80 =25 Y=mX+c y 2 x =25( 1 x )4 y= 254x


SPM Practice Question 10 – 12

Question 10:
The sequence –11, –5, 1,… is an arithmetic progression. State the three consecutive terms of this arithmetic progression where the sum of these three terms is 93.

Solution:



Question 11:
An arithmetic series, with the first term 12 and common difference d, consists of 23 terms. Given that the sum of the last 3 terms is 5 times the sum of the first 3 terms, find
(a) the value of d,
(b) the sum of the first 19 terms.

Solution:




Question 12:
The sum of n terms of an arithmetic progression is given by the formula S n = n 2 (53n) . Find
(a) the first term,
(b) the common difference,
(c) the tenth term.

Solution:


SPM Practice Question 7 – 9

Question 7:
Find the sum of all the multiples of 7 between 100 and 500.

Solution:




Question 8:
If log 10 p,  log 10 pq and  log 10 p q 2 are the first three terms of a progression, show that it forms an arithmetic progression.

Solution:




Question 9:


Show that the volumes of the cylinders in the above diagram form an arithmetic progression and state its common difference.

Solution:


SPM Practice Question 1 – 3


Question 1:
The third and eighth terms of an arithmetic progression are –5 and 15 respectively. Find
(a) The first term and the common difference
(b) The sum of the first 10 terms.

Solution:
(a)
T 3 =5    Use  T n =a+( n1 )d a+2d=5( 1 ) T 8 =15 a+7d=15( 2 ) ( 2 )( 1 ),  5d=20 d=4 Substitute d=4 into ( 1 ), a+2( 4 )=5    a=13

(b)
S 10 = 10 2 [ 2( 13 )+9( 4 ) ]Use  S n = n 2 [ 2a+( n1 )d ] =50


Question 2:
The first three terms of an arithmetic progression are 2k, 3k + 3, 5k + 1. Find
(a) the value of k,
(b) the sum of the first 15 terms of the progression.

Solution:




Question 3:
Given an arithmetic progression p + 9, 2p + 10, 7p – 1,… where p is a constant. Find
(a) value of p,
(b) the sum of the next five terms.

Solution:


The nth Term of Arithmetic Progression (Example 1 & 2)


The nth Term of Arithmetic Progression (Examples)
 
Example 1:
If the 20th term of an arithmetic progression is 14 and the 40th term is 6, 
Find
(a) the first term and the common difference,
(b) the 10th term.
 
Solution:
(a)
T20 = 14
a + 19d = 14 ----- (1) (Tn = a + (n– 1) d
T40 = – 6
a + 39d = – 6 ----- (2)
 
(2) – (1),
20d = – 20 
d = – 1 
 
Substitute d = – 1 into (1),
a + 19 (– 1) = 14
a = 33

(b)
T10 = a + 9d
T10 = 33 + 9 (– 1)
T10 = 24



Example 2:
The 3rd term and the 7th term of an arithmetic progression are 20 and 12 respectively.
(a) Calculate the 20th term.
(b) Find the term whose value is 34.

Solution:
(a)
T3 = 20
a + 2d = 20 ----- (1) ← (Tn= a + (n – 1) d
T7 = 12
a + 6d = 12 ----- (2)
 
(2) – (1),
4d = – 8 
d = – 2 
 
Substitute d = – 2 into (1),
a + 2 (– 2) = 20
a = 24
T20 = a + 19d
T20 = 24 + 19 (– 2)
T20 = –4


(b)
Tn = –34
a + (n – 1) d = –34
24 + (n – 1) (–2) = –34
(n – 1) (–2) = –58
n – 1 = 29
n = 30

(C) The nth term of an Arithmetic Progression

1.2.2 The nth term of an Arithmetic Progression

(C) The nth term of an Arithmetic Progression
 
Tn = a + (n − 1) d
where
a = first term
d = common difference
n = the number of term
Tn  = the nth term



(D) The Number of Terms in an Arithmetic Progression
Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 1
:
Find the number of terms for each of the following arithmetic progressions.
(a) 5, 9, 13, 17... , 121
(b) 1, 1.25, 1.5, 1.75,..., 8

Solution:
(a)
5, 9, 13, 17... , 121
AP,
a = 5, d = 9 – 5 = 4
The last term, Tn = 121
a + (n – 1) d = 121
5 + (n – 1) (4) = 121
(n – 1) (4) = 116
(n – 1) = 116 4  = 29
n = 30

(b)
1, 1.25, 1.5, 1.75,..., 8
AP,
a = 1, d = 1.25 – 1 = 0.25
Tn = 8
a + (n – 1) d = 8
1 + (n – 1) (0.25) = 8
(n – 1) (0.25) = 7
(n – 1) = 28
n = 29


(E) The Consecutive Terms of an Arithmetic Progression
 
If a, b, c are three consecutive terms of an arithmetic progression, then
cb = b a

Example 2:
If x + 1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression, find the value of x and its common difference.

Solution:
x + 1, 2x + 3, 6
cb = a
6 – (2x + 3) = (2x + 3) – (x + 1)
6 – 2x – 3 = 2x + 3 – x – 1
3 – 2x = x + 2
x = 1 3 1 3 + 1 , 2 ( 1 3 ) + 3 , 6 4 3 , 3 2 3 , 6 d = 3 2 3 4 3 = 2 1 3

1. Arithmetic progression

Progression

1.2.1 Arithmetic progression
(A) Characteristics of Arithmetic Progression
An arithmetic progression is a progression in which the difference between any term and the immediate term before is a constant.  The constant is called the common difference, d. 

d = Tn – Tn-1   or  d = Tn+1 – Tn

Example 1:
 
Determine whether the following number sequences is an arithmetic progression (AP) or not.
(a) –5, –3, –1, 1, …
(b) 10, 7, 4, 1, -2, …
(c) 2, 8, 15, 23, …
(d) 3, 6, 12, 24, …
 
Smart TIPS: For an arithmetic progression, you always plus or minus a fixed number

Solution:



(B) The steps to prove whether a given number sequence is an arithmetic progression

Step 1: List down any three consecutive terms. [Example: T1 , T2 , T3 .]
Step 2: Calculate the values of T3  T2 and T2  T1 .
Step 3: If T3  T2 = T2  T1 = d, then the number sequence is an arithmetic progression.
[Try Question 8 and 9 in SPM Practice 1 (Arithmetic Progression)]

Example 2:
 
Prove whether the following number sequence is an arithmetic progression
(a) 7, 10, 13, …
(b) –20, –15, –9, …

Solution:
(a)
7, 10, 13 ← (Step 1: List down T1 , T2 , T3 )
T3 T2 = 13 – 10 = 3(Step 2: Find T3 T2 and T2 T1)
T2 T1 = 10 – 7 = 3(Step 2: Find T3 T2 and T2 T1)
T3 T2 = T2 T1
Therefore, this is an arithmetic progression.

(b)
 –20, –15, –9
T3 T2 = –9 – (–15) = 6
T2 T1 = –15 – (–20) = 5
T3 T2T2 T1
Therefore, this is not an arithmetic progression.