# Solving Geometric Sequences

Updated on August 7, 2018 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

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