1, a2, a3, ..., a2n - 1, b are in AP, a,..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> 1, a2, a3, ..., a2n - 1, b are in AP, a,..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> 1, a2, a3, ..., a2n - 1, b are in AP, a,..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />
If a, a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..., a<sub>2n - 1</sub>, b are in AP, a, b<sub>1</sub>, b<sub>2</sub>, ... b<sub>2n - 1</sub>, b are in GP and a, c<sub>1</sub>, c<sub>2</sub>, c<sub>3</sub>, ..., c<sub>2n - 1</sub>, b re in HP, where a, b are positive, then the equation a<sub>n</sub>x<sup>2</sup> - b<sub>n</sub>x + c<sub>n</sub> = 0 has its roots
\(a_n\), \(b_n\), \(c_n\) are the middle terms of the AP, GP and HP between a and b. So they are the AM, GM and HM of a and b:
\(a_n = A\), \(b_n = G\), \(c_n = H\), with \(G^2 = AH\).
Discriminant = \(b_n^2 - 4a_nc_n = G^2 - 4AH = G^2 - 4G^2 = -3G^2 < 0\)
So the roots are imaginary.
Answer: (C)
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