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