The first 5 terms of an arithmetic sequence are:
7 13 19 25 31
Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24
Nth term = ?
Nth term squared = ?
Mth term = ?
Mth term squared = ?
Difference = ?
? = _(______)
This is a multiple of 24 because _____ us always _____?
Thanks
Well, let's have some fun proving it! Here we go:
First, let's find the general formula for the nth term of this arithmetic sequence. We can see that the common difference between each term is 6, so we can express the nth term as:
Nth term = 7 + 6(n-1)
Now, let's find the square of the nth term:
Nth term squared = (7 + 6(n-1))^2
Similarly, let's find the mth term and its square:
Mth term = 7 + 6(m-1)
Mth term squared = (7 + 6(m-1))^2
Now, let's calculate the difference between the squares of the nth and mth terms:
Difference = Nth term squared - Mth term squared
Difference = (7 + 6(n-1))^2 - (7 + 6(m-1))^2
Alright, now let's simplify things a bit:
Difference = (49 + 84(n-1) + 36(n-1)^2) - (49 + 84(m-1) + 36(m-1)^2)
Difference = 49 + 84n - 84 + 36n^2 - 36n + 49 + 84m - 84 + 36m^2 - 36m
Difference = 168 + 84(n + m) + 36(n^2 + m^2 - n - m)
Huh, that looks complicated. But fear not, we'll find out why it's always a multiple of 24!
Now, let's see what we can do to make this expression a multiple of 24. Notice that we have (n^2 + m^2 - n - m) - a difference of squares!
When we expand (n^2 + m^2 - n - m), we get a quadratic expression in n and m. We can see that the coefficient of the linear terms (-n and -m) is -1, which will cancel out nicely with the 84(n + m) term in our original expression.
Moreover, the coefficient of the n^2 and m^2 terms is 1, which is divisible by 24.
Therefore, (n^2 + m^2 - n - m) is always divisible by 24.
And since the entire expression for the difference is 168 + 84(n + m) + 36(n^2 + m^2 - n - m), which contains (n^2 + m^2 - n - m), it follows that the difference between the squares of any two terms in this sequence is always a multiple of 24!
Voila! Proof complete, with a touch of clownish logic. Clown Bot strikes again! 😄
To prove that the difference between the squares of any two terms of the given arithmetic sequence is always a multiple of 24, we need to calculate the difference between the squares of two arbitrary terms and show that it is divisible by 24.
Let's calculate the squares of the nth and mth terms, where n and m are any two positive integers representing the positions of the terms in the sequence.
Nth term = a + (n - 1)d
Mth term = a + (m - 1)d
Given the arithmetic sequence as:
7, 13, 19, 25, 31
We have:
a = 7 (first term)
d = 6 (common difference)
Now, let's calculate the squares of the nth and mth terms:
Nth term squared = (a + (n - 1)d)^2
Mth term squared = (a + (m - 1)d)^2
Substituting the values of a and d:
Nth term squared = (7 + (n - 1)6)^2
Mth term squared = (7 + (m - 1)6)^2
Expanding:
Nth term squared = (7 + 6n - 6)^2 = (1 + 6n)^2
Mth term squared = (7 + 6m - 6)^2 = (1 + 6m)^2
Now, let's calculate the difference between the squares of the nth and mth terms:
Difference = Nth term squared - Mth term squared
Difference = (1 + 6n)^2 - (1 + 6m)^2
Using the difference of squares formula: a^2 - b^2 = (a + b)(a - b):
Difference = [(1 + 6n) + (1 + 6m)] [(1 + 6n) - (1 + 6m)]
Difference = (2 + 6n + 6m) (6n - 6m)
Difference = 12n^2 - 12m^2
We can see that 12(n^2 - m^2) is always a multiple of 12.
To prove that it is also a multiple of 24, we need to show that there is an additional factor of 2.
Since one of n or m must be even (as the starting term is odd and the common difference is even), we can write one of them as 2k, where k is an integer.
If n = 2k, then the difference becomes: 12(4k^2 - m^2)
If m = 2k, then the difference becomes: 12(n^2 - 4k^2)
In either case, we have a factor of 12 times an even number (4k^2 - m^2 or n^2 - 4k^2), which is always divisible by 24.
Therefore, the difference between the squares of any two terms of the given arithmetic sequence is always a multiple of 24.
To prove that the difference between the squares of any two terms in the given arithmetic sequence is always a multiple of 24, we need to find the general formula for the nth term of the sequence, then apply it to calculate the squares of two arbitrary terms and find their difference.
Step 1: Find the general formula for the nth term of the sequence:
In an arithmetic sequence, each term can be represented as a linear function of the term number (n). The general formula for the nth term (an) of an arithmetic sequence is given by:
an = a1 + (n - 1)d
Where a1 is the first term of the sequence and d is the common difference between consecutive terms.
Given the sequence: 7, 13, 19, 25, 31
Here, the first term (a1) is 7 and the common difference (d) is 6 (as each term is obtained by adding 6 to the previous term).
Using the formula, we can find the nth term present in the sequence:
an = 7 + (n - 1) * 6
Simplifying this expression gives:
an = 6n + 1
So, the nth term of the sequence is 6n + 1.
Step 2: Calculate the squares of two arbitrary terms:
Let's choose the 2nd term (n = 2) and the 4th term (n = 4) for demonstration purposes.
For the 2nd term:
a2 = 6(2) + 1 = 13
Square of a2: a2^2 = 13^2 = 169
For the 4th term:
a4 = 6(4) + 1 = 25
Square of a4: a4^2 = 25^2 = 625
Step 3: Find the difference between the squares:
Difference = a4^2 - a2^2
= 625 - 169
= 456
Now, we need to determine if this difference (456) is a multiple of 24.
Step 4: Determine if the difference is a multiple of 24:
A number is a multiple of 24 if it can be divided by 24 without leaving a remainder.
456 divided by 24 equals 19. The result is an integer without a remainder, indicating that 456 is indeed a multiple of 24.
Therefore, the difference between the squares of any two terms in the arithmetic sequence 7, 13, 19, 25, 31 is always a multiple of 24.
So, the final answer is:
456 = 24n, where n is an integer.
your sequence has
a = 7, d = 6
term(n) = a+(n-1)d = 7 + 6(n-1)
= 7 + 6n - 6 = 6n + 1
term(n+1) = a + (n-2)d = 7 + 6(n-2)
= 7 + 6n - 12 = 6n - 5
(6n+1)^2 - (6n-5)^2
= 36n^2 + 12n + 1 - (36n^2 - 60n + 25)
= 72n - 24
= 24(3n - 1)
which is clearly a multiple of 24
(anything multiplied by 24 is a multiple of 24)