SOLVING GEOMETRIC SEQUENCES
Arithmetic sequences are formed by addition, whereas geometric sequences are formed by multiplication. Geometric sequences are also called geometric progressions .
A geometric sequence is one in which each term is multiplied by the same number to get the next term. This number is known as the common ratio r,
where r * An = An + 1 for n = 1, 2, 3, ……….
r maybe positive or negative.
Problem One :
Verify whether each of the following sequences is actually a geometric sequence.
(a) 5,000 20,000 80,000 32,0000
Here A = 5000 , r = 20,000/5,000 = 4
(b) 10, 000 5,000 2,500 1,250
A = 10,000 r = 5,000/10,000 = ½
(c ) 1,000 2,200 4,840 10,648
A = 1000 r = 2,200/1,000 = 2.2
All of the above are examples of geometric sequences.
The formula for finding the nth term of a certain geometric progression is given as :
An = A r n-1
Where A = first term
r = common ratio
r = A2/A1 = A3/A2 = An + 1/An
n = number of terms
An = the nth term
Problem Number Two :
Find the eighth term of the geometric sequence which begins with ¾ and 3/5.
Solution :
The ratio is : r = 3/5 ÷ ¾ = 4/5
A = ¾ n = 8 n – 1 = 8 – 1 = 7
Substituting to the formula above :
A 8 = ¾ * (4/5) 7
= ¾ * 16,384/78,125
= [(4,096 ) *3 ]/ 78, 125
= 12,288/78,125
The formula for finding the sum (Sn) of the first n terms of a geometric sequence with first term A and common ratio r, where r should not be equal to 1 is given as :
Sn = [ A (r n – 1 ) ]/ r – 1
Problem Number Three :
Find the sum of the first ten terms of the geometric series starting with -5 and 15.
Solution :
r = 15/-5 = -3
A = -5
n = 10
Sn = [ -5 ( -3 10 – 1)] / -3-1
= [ -5 ( 59,049 – 1)] /-4 =-5(59,048)/-4 = -295,240/-4 = 73,810
Alternative formula for Sn :
Sn = ( A – r An ) / 1 – r
Problem Number Four :
The first term of a geometric sequence is 5 and the fourth term is -320.
Find the eighth term and the sum of the first eight terms.
Solution :
We are given with A = 5, if we first use n = 4 in the formula
An = A r n – 1 we obtain,
-320 = 5 r 3
r 3 = – 320/5 = -64
r = – 4
We next use n = 8 in the formula for An and Sn
A 8 = 5 ( -4 ) 7 = 5 (-16, 384) = -81, 920
S 8 = (A – r An )/ 1 – r
= [ 5 – (-4) (-81,920)] /1 – (-4)
= [5 – 327, 680 ] /5
= -327, 675 / 5
= -65, 535
Problem Number Five:
Find r and A if S5 = 1, 563 and A5 = 1, 875
Solution :
First, we use the formula for An :
1,875 = A r 4 let this be equation (1 ).
Then we use the formula for Sn,
1, 563 = (A – 1,875 r) / 1 – r let this be equation (2)
Solving the second equation for A we obtain,
( 1 – r ) (1,563) = A – 1, 875 r
1,563 – 1,563r = A – 1, 875 r
312 r + 1,563 = A or A = 312 r + 1, 563, let this be equation (3)
Substituting this value in the first equation, we now have
1,875 = (312 r + 1,563) r 4
1, 875 = 312 r 5 + 1,563 r 4 0r 312 r 5 + 1, 563 r 4 – 1, 875 = 0
By using the theorem on rational zeros of polynomial function, we find one of the
Solution to be r = -5.
Substituting r = -5 in the equation (3)
A = 1, 563 + (312) ( -5)
A = – 1,560 + 1, 563 = 3
SOURCE : COLLEGE ALGEBRA
Paul K. Rees
Fred W. Sparks Charles Sparks Rees
