In arithmetic sequences, consecutive numbers are separated by a constant.
In geometric sequences, consecutive numbers are separated by a factor.
Essentially, arithmetic and geometric sequences are like linear and exponential functions, respectively, but only defined for natural numbers. Examining the equations, we can classify the functions.
Linear:& t(n)=50-7n
Exponential:& h(n)=4* 3^n
Quadratic:& q(n)=n^2-6n+17
As we can see, the first function is arithmetic, the second is geometric and the last one is neither.
b If the functions have a term in common, there exists a whole number value of n where t(n), h(n), and q(n) are all equal. Let's start by calculating some values for t(n) and h(n) and look for common values.
|c|c|c|c|c|
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n & -5pt 50-7n -5pt & t(n) & -7pt 4* 3^n -7pt & -7pt h(n) -7pt [0.2em]
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1 & 50-7( 1) & 43 & 4* 3^1 & 12 [0.2em]
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2 & 50-7( 2) & 36 ✓ & 4* 3^2 & 36 ✓ [0.2em]
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3 & 50-7( 3) & 29 & 4* 3^2 & 108 [0.2em]
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4 & 50-7( 4) & 22 & 4* 3^4 & 324 [0.2em]
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5 & 50-7( 5) & 15 & 4* 3^5 & 972 [0.2em]
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6 & 50-7( 6) & 8 & 4* 3^6 & 2916 [0.2em]
When n=2, both t(n) and h(n) equal 36. Since the linear function is decreasing and the exponential function is increasing, there cannot be any more points of intersection between these functions. Therefore, n=2 is our only viable candidate. Let's test it in the quadratic function.
At n=2, the parabola has a value of 9. Therefore, there is no term that the three sequences have in common.
c From Part B, we know that t(n) and h(n) have the second term in common. Let's calculate the values for q(n) for the same input as we used in Part B.
|c|c|c|
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n & n^2-6n+17 & q(n) [0.2em]
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1 & ( 1)^2-6( 1)+17 & 12 [0.2em]
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2 & ( 2)^2-6( 2)+17 & 9 [0.2em]
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3 & ( 3)^2-6( 3)+17 & 8 [0.2em]
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4 & ( 4)^2-6( 4)+17 & 9 [0.2em]
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5 & ( 5)^2-6( 5)+17 & 12 [0.2em]
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6 & ( 6)^2-6( 6)+17 & 17 [0.2em]
Now we add the last column to the table of values from Part B and look for common values.
|c|c|c|c|
[-1em]
n & t(n) & h(n) & q(n) [0.2em]
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1 & 43 & 12 ✓ & 12 ✓ [0.2em]
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2 & 36 ✓ & 36 ✓ & 9 [0.2em]
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3 & 29 & 108 & 8 [0.2em]
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4 & 22 & 324 & 9 [0.2em]
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5 & 15 & 972 & 12 [0.2em]
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6 & 8 & 2916 & 17 [0.2em]
We see that h(1)=q(1) and t(2)=h(2).
Extra
What about values after n=6?
Luckily, we do not actually have to bother with calculating more values after t=6. This is because the linear function is at this point below the parabola and continues to decrease while the parabola is increasing. Also, the exponential function is growing at such a fast rate that the parabola will never catch up to it.
