When to use these formulas?

Question

Arithmetic:

A. a_n=a₁+(n-1)d

 

B. S_n=n(a₁+a_n)/2

 

Geometric:

C. a_n=a¹(r)^n-1

 

D. S_n=a¹(1-rⁿ)/1-r

 

Infinite Geometric:

E. S_∞ = a₁/1-r

 

Can you give me examples of problems that will require said formulas?

Answer 1

For an arithmetic sequence each subsequent term is found by adding a constant to the previous term

This constant is called the difference or 'd'

 

The formula a_n = a₁ + (n-1)d indicates that we added the difference d to the original term (n-1) times

 

Example:  Find the 10th term of sequence    4, 7, 10, 13, 16, ... 

We see that d = 3 because each new term is found by

adding 3 to the previous term.

To find the 10th term in this sequence we would have a₁ = 4, d = 3, n-1 = 9

a₁₀ = 4 + (10-1)(3) = 4 + (9)(3) = 31

 

********************************

S_n=n(a₁+a_n)/2

 

This formula is for finding the sum of the first n terms of an arithmetic sequence.

This means we add a₁ + a₂ + a₃ + ... + a_n

 

For example:

Find the sum of the first 12 terms of the arithmetic sequence a_n = 2n + 4

First we determine this is an arithmetic sequence and find a₁ and d so we can

use the formula a_n = a₁ + (n-1)d to find a₁₂

 

a₁ = 2(1) + 4 = 6

a₂ = 2(2) + 4 = 8

a₃ = 2(3) + 4 = 10

We see that this is an arithmetic sequence where d = 2

We can find a₁₂ from the formula  a_n = a₁ + (n-1)d

a₁₂ = 6 + (12-1)(2) = 6 + 22 = 28

 

We now have a₁ = 6,  a₁₂ = 28, n = 12 so we plug into S_n=n(a₁+a_n)/2

S₁₂ = 12(6+28)/2 = 12(34)/2 = 6(34) = 204

 

This means that the first twelve terms of this arithmetic sequence add to 204

*************************************

 

a_n=a₁(r)^n-1

 

A geometric sequence is where each subsequent term is found by multiplying the

previous term by a set number called the common ratio...,    a_n+1 = r(a_n)

For example  2, 6, 18, 54, 162... 

In this geometric sequence r=3 because each subsequent term is found by

multiplying the previous term by 3 .. 

 

Example:  Find the 11th term of the geometric sequence 2, 6, 18, 54, ...

As already stated r=3    a₁ = 2     n = 11

a₁₁ = 2(3)^11-1 = 2(3)¹⁰ = 118098

*********************************

Sn=a₁(1-rⁿ)/(1-r)

 

This geometric series is basically telling you how to find the sum of the

first n terms of a geometric sequence.

 

Example:  Let's look at the previously used geometric sequence   2, 6, 18, 54, 162...  

 

Find the sum of the first 8 terms of the geometric sequence 2, 6, 18, 54, 162...  

 

We have:  a₁ = 2    r = 3   n = 8 so plug into the formula

S₈ = 2(1-(3⁸))/(1-3) = 2(1-3⁸)/(-2) = -1(1-6561) = -1(-6560) = 6560

Thus the sum of the first 8 terms of the geometric sequence is 6560

**********************************

S_∞ = a₁/(1-r)

 

This is the formula to find out what the sum of a geometric sequence approaches as

the number of entries in the sequence approaches infinity.

This sum will exist if the common ratio r is in the range   -1 < r < 1 

Or  absolute value of r is less than 1.    This basically means that the

common ratio is a fraction less than 1 and greater than -1

 

Example:  Find S_∞ [4(1/2)^n-1]

This is set up in the format of a geometric sequence  a_n = a₁(r)^n-1

We have  a₁ = 4    r = 1/2

Since r is a fraction between -1 and 1 the sum exists so plug into the formula

 

S_∞ = 4/(1-(1/2)) = 4/(1/2) = 4(2) = 8

 

This means that the sum of the terms in this geometric sequence will approach 8

as the number of terms approaches infinity