Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

Question

Let a,b be two non-zero real numbers. If p and r are then roots of the equation x^2 - 8ax + 2a = 0 and q and s aren the roots of the equationn nnx^2 + 12bx + 6b = 0, such thatn(1 / p),(1 / q),(1 / r),(1 / s) are in A.P., thenna^- 1 - b^- 1 is equal to

DivineJEE · Accepted Answer

p and r are the roots of the equation,nnnnx^2-8 a x+2 a=0nnnnnnTherefore p+r=8 a, p r=2 a => (1 / p)+(1 / r)=4 \ (i)nnn nAlso, q and s are the roots of x^2+12 b x+6 b=0nnnnTherefore q+s=-12 b, q s=6 b => (1 / q)+(1 / s)=-2 \ (i i)nnn nGiven that (1 / p), (1 / q), (1 / r), (1 / s) are in A.P.nnnnTherefore (1 / q)-(1 / p)=(1 / r)-(1 / q) => (2 / q)=(1 / r)+(1 / p)=4nnnn[Using (i)]nnnn => q=(1 / 2) and s=(-1 / 4)nnn nClearly, (1 / r)-(1 / q)=(1 / s)-(1 / r) => (2 / r)=(1 / q)+(1 / s)nnnn => (2 / r)=-2 => r=-1 and p=(1 / 5)nnnn[Using (ii)]nnnnTherefore (1 / a)-(1 / b)=(2 / p r)-(6 / q s)=38nnn

Explanation

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