Let a, b, c be in G.P with common ratio r, where a≠0 and 0 < r ≤ 1/2. If 3a, 7b, 15c are the first three terms of an A.P., then the 4th term of this A.P is<p></p>

b=ar; c=ar² ⇒ 14b = 3a + 15c ⇒ 14(ar) = 3a + 15ar² ⇒ 14r = 3 + 15r² ⇒ 15r² - 14r + 3 = 0 ⇒ (3r - 1)(5r - 3) = 0 ⇒ r = 1/3, 3/5
Only acceptable value is r = 1/3, because r ∈ (0, 1/2]
∴ d = 7b - 3a = 7(ar) - 3a = 7(1/3)a - 3a = -2/3a
4th term = 15c - 2/3a = 15(1/9)a - 2/3a = a

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Question:

Let a, b, c be in G.P with common ratio r, where a≠0 and 0 < r ≤ 1/2. If 3a, 7b, 15c are the first three terms of an A.P., then the 4th term of this A.P is

Solution: