c The equations describing arithmetic and geometric sequences are linear and exponential, respectively. Find an equation that is neither.
A
a 9,15,21,27
Equation: t(n)=6n-3
B
b 9,27,81,243
Equation: t(n)=3(3)^(n-1)
C
cExample Solution: 12,27,48,75
Equation: t(n)=3n^2
Practice makes perfect
a For a sequence to be arithmetic sequence, the difference between consecutive term must be constant. The difference between 3 and 9 is 6, so to find the third term we add 6 to 9.
To find the following three terms, we just continue this pattern.
An arithmetic sequence can be written in the following format.
&t(n)=d(n-1) + t(1)
&d=common difference
&t(1)=first term
Let's substitute the common difference and first term into the equation.
b For a sequence to be geometric, there must be a factor separating consecutive terms. The ratio of 9 to 3 is 3, so to find the third term we must multiply the second term by 3.
To find the following three terms, we just continue this pattern.
A geometric sequence and can be written in the following format.
&t(n)=t(1)r^(n-1) [0.3em]
&r=common factor
&t(1)=first term
We know that the common factor is r= 3 and the first term is t(1)= 3. With this information, we can write the equation.
t(n)= 3( 3)^(n-1)
c We are asked to find a sequence that starts with 3, meaning t(1)=3, but that is neither arithmetic nor geometric. Therefore, the equation describing the sequence cannot be linear or exponential. But it can, for example, be quadratic.
t(n)=k(n)^2
The first term is 3, so t(1)=3. We use this relation to determine the coefficient k.
The coefficient k is 3, so the sequence can be described by the following equation.
t(n)=3(n)^2
This is neither arithmetic nor geometric. To find the next four terms we substitute 2,3,4, and 5 for n in our formula.
|c|c|c|
[-1em]
n & 3(n)^2 & t(n) [0.2em]
[-1em]
2 & 3( 2)^2 & 12 [0.2em]
[-1em]
3 & 3( 3)^2 & 27 [0.2em]
[-1em]
4 & 3( 4)^2 & 48 [0.2em]
[-1em]
5 & 3( 5)^2 & 75 [0.2em]
The next four terms in the sequence are 12, 27, 48, 75.
