A quadratic sequence begins:
[tex]$5,20,45,80,125, \ldots$[/tex]
Work out the [tex]$n^{\text {th }}$[/tex] term rule for this sequence.
Answer (2)
Calculate the first differences: 15 , 25 , 35 , 45 .
Calculate the second differences: 10 , 10 , 10 .
Determine that the n t h term is of the form a n 2 + bn + c , where 2 a = 10 , so a = 5 .
Solve for b and c using the first two terms of the sequence, resulting in b = 0 and c = 0 . The n t h term rule is 5 n 2 .
Explanation
Problem Analysis We are given the quadratic sequence 5 , 20 , 45 , 80 , 125 , … and we need to find the n t h term rule for this sequence.
First Differences Let's find the first differences between consecutive terms: 20 − 5 = 15 , 45 − 20 = 25 , 80 − 45 = 35 , 125 − 80 = 45 . The first differences are 15 , 25 , 35 , 45 , …
Second Differences Now let's find the second differences between consecutive terms of the first differences: 25 − 15 = 10 , 35 − 25 = 10 , 45 − 35 = 10 . The second differences are constant and equal to 10 .
Finding the coefficient a Since the second differences are constant, the n t h term of the sequence can be expressed in the form a n 2 + bn + c . The second difference is equal to 2 a , so 2 a = 10 , which means a = 5 .
Setting up the equations Now we know the n t h term is of the form 5 n 2 + bn + c . We can use the first two terms of the sequence to form a system of equations to solve for b and c . For n = 1 , the term is 5 , so 5 ( 1 ) 2 + b ( 1 ) + c = 5 , which simplifies to 5 + b + c = 5 , or b + c = 0 . For n = 2 , the term is 20 , so 5 ( 2 ) 2 + b ( 2 ) + c = 20 , which simplifies to 20 + 2 b + c = 20 , or 2 b + c = 0 .
Solving for b and c We have the system of equations:
b + c = 0 2 b + c = 0
Subtracting the first equation from the second equation, we get ( 2 b + c ) − ( b + c ) = 0 − 0 , which simplifies to b = 0 . Since b + c = 0 , we have 0 + c = 0 , so c = 0 .
The nth term rule Therefore, the n t h term rule for the sequence is 5 n 2 + 0 n + 0 = 5 n 2 .
Verification Let's check if this rule works for the first few terms: For n = 1 , 5 ( 1 ) 2 = 5 For n = 2 , 5 ( 2 ) 2 = 5 ( 4 ) = 20 For n = 3 , 5 ( 3 ) 2 = 5 ( 9 ) = 45 For n = 4 , 5 ( 4 ) 2 = 5 ( 16 ) = 80 For n = 5 , 5 ( 5 ) 2 = 5 ( 25 ) = 125 The rule works!
Final Answer The n t h term rule for the sequence 5 , 20 , 45 , 80 , 125 , … is 5 n 2 .
Examples
Understanding quadratic sequences is useful in various fields. For example, the distance an object falls under constant acceleration (like gravity) follows a quadratic pattern. If you drop a ball from a height, the distance it falls each second forms a sequence. Knowing the n t h term rule helps predict how far the ball will have fallen after any given number of seconds. This principle is applied in physics and engineering to model motion and trajectories.
To find the n th term rule of the quadratic sequence 5 , 20 , 45 , 80 , 125 , … , we calculate the first and second differences, leading to the conclusion that the rule is 5 n 2 . This has been verified against the initial sequence values, confirming its accuracy.
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