Question

If a, a₁, a₂, a₃, ..., a_{2n - 1}, b are in AP, a, b₁, b₂, ... b_{2n - 1}, b are in GP and a, c₁, c₂, c₃, ..., c_{2n - 1}, b re in HP, where a, b are positive, then the equation a_nx² - b_nx + c_n = 0 has its roots
1. Real and equal
2. Real and unequal
3. Imaginary
4. One real and one imaginary

Answer 1

Correct Answer - Option 3 : Imaginary

Concept:

If a, b, c are in AP, than \(b = \frac{{a + c}}{2}\)

If a, b, c are in GP than \(b = \sqrt {ac} \)

If a, b, c are in HP than \(b = \frac{{2ac}}{{a + c}}\)

Calculation:

assume a = 2, b = 8, n = 1

2, a_1, 8 are in AP, than \({a_1} = \frac{{2 + 8}}{2} = 5\)

2, b1, 8 are in GP, than \({b_1} = \sqrt {2 \times 8} = 4\)

2, c1, 8 are in HP, than \({c_1} = \frac{{2 \times 2 \times 8}}{{2 + 8}} = \frac{{16}}{5}\)

Here, we have to find the nature of roots of the equation anx2 - bnx + cn = 0

By substituting n = 1 in the equation anx2 - bnx + cn = 0 we get

⇒ a₁x2 - b₁x + c₁ = 0

 

We know that, a₁ = 5, b₁ = 4 and c₁ = 16/5

 

So, the discrimination \(D = {b_1}^2 - 4{a_1}{c_1}\) for the equation a1x2 - b1x + c1 = 0 is \(D = {4^2} - 4 \times 5 \times \frac{{16}}{5} = - 48\)

 

∵ D < 0 so the roots are imaginary

 

Hence, option C is the correct answer.