3. The nth Term of Geometric Progressions (Part 1)

1.4.2 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions

T n = a r n 1

a = first term
r = common ratio
n = the number of term
Tn = the nth term

Example 1:
Find the given term for each of the following geometric progressions.
(a) 8 ,4 ,2 ,...... T8
(b) 16 27 , 8 9 , 4 3 , ..... ,  T6

Solution:
T n = a r n 1 T 1 = a r 1 1 = a r 0 = a ( First term ) T 2 = a r 2 1 = a r 1 = a r ( S e c o n d term ) T 3 = a r 3 1 = a r 2 ( T h i r d term ) T 4 = a r 4 1 = a r 3 ( Fourth term )

(a)
8 , 4 , 2 , ..... a = 8 , r = 4 8 = 1 2 T 8 = a r 7 T 8 = 8 ( 1 2 ) 7 = 1 16

(b)
16 27 , 8 9 , 4 3 , ..... a = 16 27 r = T 2 T 1 = 16 27 8 9 = 2 3 T 6 = a r 5 = 16 27 ( 2 3 ) 5 = 512 6561

(F) Sum of the First n Terms of an Arithmetic Progression 

1.2.3 Sum of the First nTerms of an Arithmetic Progression 

(F) Sum of the First n terms of an Arithmetic Progressions
S n = n 2 [ 2 a + ( n 1 ) d ] S n = n 2 ( a + l )
a = first term
d = common difference
n = the number of term
Sn = the sum of first n terms


Example:
Calculate the sum of each of the following arithmetic progressions.
(a) -11, -8, -5, ... up to the first 15 terms.
(b) 8,   10½,   13,...   up to the first 13 terms.
(c) 5, 7, 9,....., 75 [Smart TIPS: The last term is given, you can find the number of term, n]
 
Solution:
(a)
11 , 8 , 5 , ..... Find S 15 a = 11 , d = 8 ( 11 ) = 3 S 15 = 15 2 [ 2 a + 14 d ] S 15 = 15 2 [ 2 ( 11 ) + 14 ( 3 ) ] = 150

(b)
8 , 10 1 2 , 13 , ..... Find S 13 a = 8 d = 10 1 2 8 = 5 2 S 13 = 13 2 [ 2 a + 12 d ] S 13 = 13 2 [ 2 ( 8 ) + 12 ( 5 2 ) ] = 299

(c)
5 , 7 , 9 , ..... , 75 ( The last term l = 75 ) a = 5 d = 7 5 = 2 S n = n 2 ( a + l ) S 36 = 36 2 ( 5 + 75 ) = 1440 The last term l = 75 T n = 75 a + ( n 1 ) d = 75 5 + ( n 1 ) ( 2 ) = 75 ( n 1 ) ( 2 ) = 70 n 1 = 35 n = 36


SPM Practice Question 1 – 3


Question 1:
The fourth term of a geometric progression is –20. The sum of the fourth and the fifth term is –16. Find the first term and the common ratio of the progression.

Solution:





Question 2:
The fourth and the seventh terms of a geometric progression are 18 and 486 respectively. Find the third term.

Solution:




Question 3:
For a geometric progression, the sum of the first two terms is 30 and the third term exceeds the first term by 15. Find the common ratio and the first term of the geometry progression.

Solution:
T 1 + T 2 =30 a+ar=30 a( 1+r )=30(1) T 3 T 1 =15 a r 2 a=15 a( r 2 1 )=15(2) ( 2 ) ( 1 ) = a( r 2 1 ) a( 1+r ) = 15 30 ( r1 )( r+1 ) 1+r = 1 2 ( r 2 1 )= ( r1 )( r+1 ) r1= 1 2 r= 1 2 +1= 3 2 From (1),   a( 1+r )=30 a( 1+ 3 2 )=30   5a 2 =30 a=12

The nth Term of Geometric Progression (Examples)


Example 1:
The sixth term of a geometric progression is 32 and the third term is 4. Find the first term and the common ratio.
Smart TIPSSolving the simultaneous equation of a and rUsing the formula T n = a r n−1

Solution:
T 6 =32 a r 5 =32 ----- (1) T 3 =4 a r 2 =4 ----- (2) (1) (2) = a r 5 a r 2 = 32 4 r 3 =8 r=2 Substitute r=2 into (2), a ( 2 ) 2 =4 a=1



Example 2:
In a geometric progression, the sum of the second term and the third term is 12 and the sum of the third term and the fourth term is 4, find the first term and the common ratio.
[Smart TIPS: Solving simultaneous equation to find a and r]

Solution:
T2 + T3 = 12
ar + ar2  = 12
ar (1 + r) = 12 ----- (1) ← (Factorisation)
T3 + T4 = 4
ar2 + ar3  = 4
  ar2 (1 + r) = 4 ----- (2)

(2) (1) = a r 2 ( 1+r ) ar( 1+r ) = 4 12 r= 1 3 Substitute r= 1 3  into (1), a( 1 3 )( 1+ 1 3 )=12 a=27



Example 3
Find which term in the progression 3, 12, 48, , ... is the first to exceed 1 000 000.
[Smart TIPS: Using Tn formula for solving n]

Solution:
3 , 12 , 48 , ..... G P , a = 3 , r = T 2 T 1 = 12 3 = 4 T n > 1000000 ( 3 ) ( 4 ) n 1 > 1000000 ( 4 ) n 1 > 1000000 3 [ ( 3 ) ( 4 ) n 1 12 n 1 ] log 4 n 1 > log 1000000 3 ( Put log for both sides) ( n 1 ) lg 4 > lg 1000000 3 ( log a m n = n log a m ) ( n 1 ) ( 0.6021 ) > 5.523 n 1 > 9.17 n > 10.17 n = 11 ( n is an integer) C h e c k : T 11 = ( 3 ) ( 4 ) 10 T 11 = 3145728 > 1000000

SPM Practice Question 1


Question 1:
Diagram below shows part of an arrangement of bricks of equal size.



The number of bricks in the lowest row is 100. For each of the other rows, the number of bricks is 2 less than in the row below. The height of each brick is 7 cm.

Rahman builds a wall by arranging bricks in this way. The number of bricks in the highest row is 4. Calculate

(a) the height, in cm, of the wall.
(b) the total price of the bricks used if the price of one brick is 50 sen.

Solution:
100, 98, 96, …, 4 is an arithmetic progression
a = 100 and d = –2

(a)
Tn = 4
a + (n – 1) d = 4
100 + (n – 1)(–2) = 4
100 – 2n + 2 = 4
2n = 98
n = 49
Height of the wall = 49 × 7 = 343 cm

(b)
Total number of bricks used
= S49
= 49 2 ( 100+4 ) S n = n 2 ( a+l )
= 2548
Total price = 2548 × RM0.50
= RM1,274

SPM Practice Question 10 – 12


Question 10:
In a geometric progression, the first term is 27 and the fourth term is 1. Calculate
(a) the positive value of common ratio, r,
(b) the sum of the first n terms where n is sufficiently large till r n 0

Solution:





Question 11:
Express the recurring decimal 0.187187187 … as a fraction in its simplest form.

Solution:





Question 12:
Given that  1 p =0.1666666.....  =q+a+b+c+.....
where p is a positive integer. If q = 0.1 and a + b + c are the first three terms of a geometric progression, state the value of a, b and c in decimal form. Hence find the value of p.

Solution:


SPM Practice Question 1


Question 1:
Diagram below shows part of an arrangement of bricks of equal size.



The number of bricks in the lowest row is 100. For each of the other rows, the number of bricks is 2 less than in the row below. The height of each brick is 7 cm.

Rahman builds a wall by arranging bricks in this way. The number of bricks in the highest row is 4. Calculate

(a) the height, in cm, of the wall.
(b) the total price of the bricks used if the price of one brick is 50 sen.


Solution:
100, 98, 96, …, 4 is an arithmetic progression
a = 100 and d = –2

(a)
Tn = 4
a + (n – 1) d = 4
100 + (n – 1)(–2) = 4
100 – 2n + 2 = 4
2n = 98
n = 49
Height of the wall = 49 × 7 = 343 cm

(b)
Total number of bricks used
= S49
= 49 2 ( 100+4 ) S n = n 2 ( a+l )
= 2548
Total price = 2548 × RM0.50
= RM1,274

2. Geometric progression


2. Geometric Progression
(A) Characteristics of Geometric Progressions 
Geometric progression is a progression in which the ratio of any term to the immediate term before is a constant. The constant is called common ratio, r.

Example:
Determine whether or not each of the following number sequences is a geometric progression (GP).
(a) 1, 4, 16, 64, …..
(b) 10, –5, 2.5, –1.25, …..
(c) 2, 4, 12, 48, …..
(d) –6, 1, 8, 15, …..

Solution:
[Smart Tips: For GP, times a fixed number every time to get the next time.]

(a)

Common ratio, r = T n T n 1 T 3 T 2 = 16 4 = 4 , T 2 T 1 = 4 1 = 4 T 3 T 2 = T 2 T 1
This is a GP, a =1, r = 4.

(b)

Common ratio, r = T 2 T 1 = 5 10 = 1 2
This is a GP, a =1, r = – ½.

(c)

This is NOT a GP.
This is because the ratio of each term to its preceding term is not a similar constant.



(B) The steps to prove whether a given number sequence is a geometric progression.
Step 1: List down any three consecutive terms. [Example: T1, T2, T3.]
Step 2: Calculate the values of  T 3 T 2 and T 2 T 1 .    
Step 3: If  T 3 T 2 = T 2 T 1 = r , then the number sequence is a geometric 
 progression.
Step 4: If  T 3 T 2 T 2 T 1 , then the number sequence is not a geometric progression.

(F) Sum of the First n Terms of Geometric Progressions

1.4.3 Sum of the First n Terms of Geometric Progressions

(F) Sum of the First Terms of Geometric Progressions
S n = a ( r n 1 ) r 1 , r > 1 S n = a ( 1 r n ) 1 r , r < 1 a = first term
r = common ratio
n = number of term
Sn = sum of the first n terms

Example 1:
Find the sum of each of the following geometric progressions.
(a) 1, 2, 4, ...  up to the first 7 terms
(b) 9, 3,   1,   ,   ...    up to the first 6 terms
(c) 12, 3, ...., 3 64 [Smart TIPS: You can find n if you know the last term]


Solution:







SPM Practice Question 7 – 9


Question 7:
Given a geometric progression 2 z , 3,  9z 2 , q,.... express q in terms of z.

Solution:





Question 8:
The second and the fourth term of a geometry progression are 10 and 2 5 respectively. Find
(a) The first term and the common ratio where r > 0,
(b) The sum to infinity of the geometry progression.

Solution:





Question 9:
In a geometric progression, the first term is 18 and the common ratio is r.
Given that the sum to infinity of this progression is 21.6, find the value of r.

Solution: