Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.
The possible GPs are -4/3, -8/3, -16/3, ... and 4, -8, 16, -32, ...
Explanation
To find the GP, we use the standard properties of a geometric progression. The first term is 'a' and the common ratio is 'r'. The sum of the first two terms gives a linear equation in 'a' and 'r'. The condition relating the fifth and third terms gives a quadratic equation in 'r', yielding two possible values for the common ratio. Substituting these values back into the linear equation gives the corresponding first terms, resulting in two valid geometric progressions.
Solution Steps
Step 1: Let the first term of the GP be a and the common ratio be r.
Step 2: The sum of the first two terms is a + ar = -4, which simplifies to a(1 + r) = -4. (Equation 1)
Step 3: The fifth term is ar⁴ and the third term is ar². Given that the fifth term is 4 times the third term, we write ar⁴ = 4ar².
Step 4: Dividing both sides by ar² (since a 0 and r 0), we get r² = 4, which means r = 2 or r = -2.
Step 5: Case 1 (r = 2): Substituting r = 2 into Equation 1 gives a(1 + 2) = -4, so 3a = -4 and a = -4/3. The GP is -4/3, -8/3, -16/3, ...
Step 6: Case 2 (r = -2): Substituting r = -2 into Equation 1 gives a(1 - 2) = -4, so -a = -4 and a = 4. The GP is 4, -8, 16, -32, ...