Geometric Progression

Objectives The presentation intends to:  teach students how to solve problems related to geometric progressions;  help students understand more about the given topic;

What are geometric progressions?  It is a sequence in which each term is obtained by multiplying the preceding term by a constant  A sequence t 1,t 2,.... t n,... is called a geometric sequence ( or more commonly a geometric progression) if there exists a number r such that t n /t n-1 = r for n>1

 Also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.

Symbols t n = last term n= number of terms r= common ratio a= first term S n = sum of terms

Formulas t n = ar n-1 a= t n /r n-1 S n = a(1-r n )/1-r r= t n /t n-1

Examples Find the common ratio and general term of Geometric Progression 1)1, 2/3, 4/9,…2) 2,-4, 8,… r= 2/3 r= -4/ 2 1 = -2 =2/3 t n = 2(-2) n-1 t n = (2/3) n-1

3) √2, √6, 3√2,… r= √6 √2 =√3 t n =√2(√3) n-1 4) If x+3, 4x, 2x+18, form geometric progression, find x. 4x = 2x+18 x+3 4x 16x 2 = 2x 2 +6x+18x+54 14x 2 -24x-54=0 2 7x 2 -12x-27=0 (7x+9)(x-3)=0 x= -9 x=3 7

5) The fifth term of a geometric progression is 81 and the ninth term is 16. Find the first term, common ratio, and the n th term. t 5 = 81, t 9 =16, a=?, r=?, n th term= ? t n =ar n-1 t 5 = ar 5-1 t 9 = t 5 r =a(2/3) 4 16= 81r /16 81 t n =ar n-1 4 √16 = 2 t n = 6561/16(2/3) n-1 4 √81 3

Insert three positive geometric means between a)1/2 and 128 ½, t 2, t 3, t 4, 128 t 5 = ar = ½ r 4 ½ 256= r 4 4 √r4=4 √256 r=4 t2=1/2(4)=2 t3=2(4)= 8 t4=8(4)= 32

t5= ar5-1 4= 1r4 4 √r4=4 √4 r= 4√22 r= 2 2/4 r= 2 ½ 1, t 2, t 3,t 4, b)1 and 4 t2= 1 √2 t3= (√2) (√2) = 2 t4= 2 (√2) = 2√2 r= √2

Find the sum of the first ten terms: … S n = a(1-r n ) 1-r S 10 = 2[1-(2) 10 ] 1-2 S10= 2(1-1624) S10= 2046

Find the sum of the geometric progression S n = a(1-r n ) 1-r S 8 = 10[1-(- 1 / 5 ) 8 ] 1-(- 1 / 5 ) S 8 = 10( ) 1.2 S 8 = , -2, 2/5,… to 8 terms

Find the t 5 of the geometric progression whose first element is -3 and whose common ratio is -2. t n = ar n-1 t 5 = (-3)(-2) 5-1 t 5 = (-3)(-2) 4 t 5 = (-3)(16) t 5 = -48

Infinite Arithmetic and Geometric Progression

Definition The sum of an infinite geometric progression a, ar, ar 2, … with ‌‌ r ‌ < 1 is given by, S=a/1-r whereas, -1<r<1 S is also called the sum to infinity of the G.P. a, ar, ar 2 … In symbols, we write this as, S=lim S n and read “S is the limit of S n as n increases without bound”

The rational number 1/3 has repeating decimal.33333… which we can consider to be as infinite sum: 1/3 = … This is an example of an infinite geometric series with ratio.1. Thus,.03 =.3x.1,.003 =.03x.1,.0003 =.0003x.1, and so on. We define an infinite geometric series in the following way.

Find the sum of the following terms. 1) … =infinite 2) … =infinite 3) 9+3+1/3… r=1/3 S= a 1-r S= 9 1-1/3 S= 9 2/3 S= 27/2 4) /2… r=1/2 S= a 1-r S= /2 S= 50 1/2 S= 100

Exercises:

A. Determine whether the sequence is a geometric progression or not. If the sequence is a geometric progression give the common ratio (r) and the next three terms. 1) 1 2,2 2,3 2 5) 3, 3√3, 9 2) 1, - 1 / 2, 1 / 3 6) 2, √3, 2 / 3 3) 1 / 6, 1 / 2, 3 / 2 7) 3.33, 2.22, ) 4, 3, 9 / 4 8) -36, -2, - 1 / 9

B. Find the indicated term of the geometric progression whose first element is a and whose common ratio is r. a= ½, r= 2/3, find t 4 a= 4, r=-1/2, find t 5 a= 81, r= 1/3, find t 7 a= 1/3, r= 3/2, find t 6

C. Find the sum to the infinity of the geometric progression , 4, 1 2. ½, 1/6, 1/ , -1/2, -1/8 4. 2/3, 1/9, 1/ , 1.212,

1. Insert five geometric means between a. 1/2 and 32 b. 2 and 54

E. For the specific value of n, find the nth term of each geometric series which starts as follows. 1. (a) …, n = 7 (b) …, n = 9 2. (a) …, n = 8 (b) …, n = 6

Solutions and Answers

A. 1) 1 2,2 2,3 2 Not geometric progression 2) 1, - 1 / 2, 1 / 3 Not geometric progression 3) 1 / 6, 1 / 2, 3 / 2 r= 1 / 2 = 3 1 / 6 ( 3 / 2 )(3)= 9 / 2 ( 9 / 2 )(3)= 27 / 2 ( 27 / 2 )(3)= 81 / 2 next terms are 9 / 2,, 27 / 2, 81 / 2 4) 4, 3, 9/4 r= 3/4 (3/4)(9/4)= 27/16 (3/4)(27/16)= 81/64 (3/4)(81/64)= 243/256 next terms are: 27/16, 81/64, 243/256

5) 3, 3√3, 9 r= √3 (√3)(9)= 9√3 (√3) (9√3)= 27 (√3)(27)=27√3 next terms are: 9√3, 27, 27√3 6) 2, √3, 2 / 3 Not geometric progression 7) 3.33, 2.22, 1.11 Not geometric progression 8) -36, -2, -1/9 r= -2/-36= 1/18 (-1/9)(1/18)= -1/162 (-1/162)(1/18)= -1/2916 (-1/2916)(1/18)= -1/52488 next terms are: -1/162, -1/2916, -1/52488

B. 1)a= ½, r= 2 / 3, t 4 = ? t n = ar n-1 t 4 = (½)( 2 / 3 ) 4-1 t 4 = (½)( 2 / 3 ) 3 t 4 = (½)( 8 / 27 ) t 4 = 4 / 27 2) a= 4, r=-1/2, t 5 = ? t n = ar n-1 t 5 = (4)(-1/2) 5-1 t 5 = (4)(-1/2) 4 t 5 = (4)(1/16) t 5 = 4/16 t 5 = ¼

3) a= 81, r= 1 / 3, t 7 =? t n = ar n-1 t 7 = (81)( 1 / 3 ) 7-1 t 7 = (81)( 1 / 3 ) 6 t 7 = (81)( 1 / 729 ) t 7 = 1 / 9 4) a= 1/3, r= 3/2, t 6 =? t n = ar n-1 t 6 = (1/3)(3/2) 6-1 t 6 = (1/3)(3/2) 5 t 6 = (1/3)(7 19/32) t 6 = 2 17/32

1) 16, 4, 1 r= 1 / 4 S= a 1-r S= / 4 S= 16 3 / 4 S= 64 / 3 2) ½, 1/6, 1/18 r= 1/3 S= a 1-r S= ½ 1-1/3 S= ½ 2/3 S= 3/4 C.

3) -2, - 1 / 2, - 1 / 8 r= ¼ S= a 1-r S= / 4 S= -2 3 / 4 S= - 8 / 3 4) 2/3, 1/9, 1/54 r= 1/6 S= a 1-r S= 2/3 1-1/6 S= 2/3 5/6 S= 4/5 5) 5.05, 1.212, r= 0.24 S= a 1-r S= S= S= 6 49/76

a. 1/2 and 32 t7=t1r 6 32= ½ r 6 ½ 64= r 6 6√64=r b. 2 and 54 t 7 = t 1 r 6 54=2r √27= 6 √ r 6 r= 6 √27 r= 6 √3 3 r=2 t 2 =1 t 3 =2 t 4 =4 t 5 =8 t 6 =16 r= √3 t 2 = 2 √3 t 3 = 6 t 4 = 6 √3 t 5 = 18 t 6 = 18 √3

E. 1. (a) …,,n=7 t n = ar n-1 t 7 = (2)(3) 6 =1458 (b) …, n=9 t n = ar n-1 t 9 = (5)(2) 8 = 1280

2. (a)4+2+1+…,n=8 t n =ar n-1 t 7 =(4)(1/2) 7 =1/32 (b) …,n=6 t n = ar n-1 t 6 =(25)(-1/5) 5 = -1/125

Prepared by: Princess Fernandez Karen Calanasan Kyla Villegas