Let the coefficients of the middle terms in the expansion of (1/√6+β x)⁴,(1-3 β x)² and (1-β/2 x)⁶, β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then 50-(2 d)/β² is equal to __________.

Solution

Maths sir · 4 steps

Find the three middle-term coefficients, then use 2b = a + c to pin down β.

1. Three middle terms. For power 4 the middle term is T₃: C(4, 2)(1/√6)²β² = 6 × 1/6β² = β². For power 2 it is T₂: C(2, 1)(-3β) = -6β. For power 6 it is T₄: C(6, 3)(-β/2)³ = 20 × (-β³/8) = -5β³/2.

   2(-6β) = β² - 5β³/2

2. Solve for β. Divide by β (allowed, since β > 0) and multiply by 2: 5β² - 2β - 24 = 0. So β = (2 ± 22)/10, and the positive root is β = 12/5.
3. Common difference. d = -6β - β², so 2d/β² = -12/β - 2 = -5 - 2 = -7.
4. So. 50 - 2d/β² = 50 - (-7) = 57.

The common slip: 43. Taking d as first term minus second gives 2d/β² = +7, and the answer comes out as 43 instead of 57.

The idea to keep. Even power n has exactly one middle term, T_(n/2+1). Keep the sign of the second term inside the bracket, and for three terms in A.P. use 2(middle) = first + third.

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