9514 1404 393
Answer:
the difference of squares is 24(3n+2), so a multiple of 24
Explanation:
The sequence has a first term of 7 and a common difference of 6, so the n-th term is ...
a[n] = a1 +d(n -1)
a[n] = 7 + 6(n -1)
a[n] = 6n +1
Then the (n+1)-th term is ...
a[n+1] = 6(n+1) +1 = 6n +7
The difference of the squares of these terms is ...
a[n+1]^2 -a[n]^2 = (6n+7)^2 -(6n+1)^2
= (36n^2 +84n +49) -(36n^2 +12n +1)
= 72n +48
= 24(3n +2) . . . . . always a multiple of 24
The difference of squares of terms n and n+1 is 24(3n+2), a multiple of 24.
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Check
The difference between squares for n=1 and n=2 is 13^2 -7^2 = 169 -49 = 120. The formula tells us the difference is 24(3·1 +2) = 24·5 = 120, in agreement.