The 1st term of an arithmetic progression is a and the common difference is d, where d ≠ 0.

(i) Write down expressions, in terms of a and d, for the 5th term and the 15th term.

The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of a geometric progression.

(ii) Show that 3a = 8d.

(iii) Find the common ratio of the geometric progression.

(i)

a_n = a + (n-1)d

a₁ = a
a₅ = a + 4d
a₁₅ = a + 14d

(ii)

Since a₁, a₅ and a₁₅ form a geometric progression, then common ratio is
r = (a + 4d) / a = (a + 14d) / (a + 4d)
a (a + 14d) = (a + 4d) (a + 4d)
a² + 14ad = a² + 8ad + 16d²
6ad − 16d² = 0
2d (3a − 8d) = 0

Since d ≠ 0, then
3a − 8d = 0
3a = 8d

(iii)

r = (a + 4d) / a

Since 8d = 3a, then 4d = 3/2 a

r = (a + 3/2 a) / a
r = (5/2 a) / a
r = 5/2