If a₁ (> 0), a₂, a₃, a₄, a₅ are in a G.P., a₂ + a₄ = 2a₃ + 1 and 3a₂ + a₃ = 2a₄, then a₂ + a₄ + 2a₅ is equal to ___________.

Solution

Maths sir · 4 steps

Write every term as arᵏ; the equation with no constant gives r, and the other one then gives a.

1. Find r first. Put a₂ = ar, a₃ = ar², a₄ = ar³ into 3a₂ + a₃ = 2a₄. Dividing by ar leaves an equation in r alone.

   2r² - r - 3 = 0 ⇒ (2r - 3)(r + 1) = 0 ⇒ r = 3/2 or r = -1

2. Reject r = -1. With r = -1, the other condition becomes -2a = 2a + 1, so a = -1/4. That breaks a₁ > 0, so r = 3/2.
3. Find a. From ar + ar³ = 2ar² + 1 we get a(12/8 + 27/8 - 36/8) = 1. So 3a/8 = 1 and a = 8/3.
4. So. a₂ + a₄ + 2a₅ = a(r + r³ + 2r⁴) = 8/3(12/8 + 27/8 + 81/8) = 8/3 × 15 = 40.

The common slip: 0. Keeping r = -1 gives a = -1/4 and a sum of 0. That value of a breaks the condition a₁ > 0, so this root has to be thrown out.

The idea to keep. In a G.P. question with two conditions, start with the one that has no constant term. The a cancels and r comes out directly. Then use the sign condition on a to choose the right root.

Answer from the NTA final answer key. Solution by PrepWiser.