Question

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

Answer 1

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)