If 3^(2 sin 2α - 1), 14 and 3^(4 - 2 sin 2α) are the first three terms of an A.P. for some α, then the sixth terms of this A.P. is:

Solution

Maths sir · 4 steps

Call 3^(2 sin 2α) = t, use the middle-term rule, and let sin 2α ≤ 1 pick the root.

1. Middle-term rule. In an A.P. the middle term is the average of its neighbours, so the first and third terms add up to 2 × 14 = 28. Put t = 3^(2 sin 2α). Then the first term is t/3 and the third term is 81/t.

   t/3 + 81/t = 28 ⇒ t² - 84t + 243 = 0 ⇒ (t-3)(t-81) = 0

2. Reject one root. t = 81 needs 2 sin 2α = 4, that is sin 2α = 2, which is impossible. So t = 3, giving sin 2α = 1/2.
3. Build the A.P. The first term is 3⁰ = 1 and the third is 3³ = 27. So the A.P. is 1, 14, 27, … with a = 1 and d = 13.
4. So. The sixth term is a + 5d = 1 + 5 × 13 = 66, option (a).

The idea to keep. When terms carry the same power of a base, replace that power with one letter. The quadratic gives two roots, so always check which ones the range of sine or cosine allows.

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