91Ó°ÊÓ

Since \(a, b, c\) are in AP, we have: \[b = \frac{a+c}{2}\] Since \(\alpha, \beta, \gamma\) are in HP, we have: \[\frac{1}{\beta} = \frac{\frac{1}{\alpha}+\frac{1}{\gamma}}{2}\] Since \(a\alpha, b\beta, c\gamma\) are in GP, we have: \[(b\beta)^2 = a\alpha \cdot c\gamma\]

From the equation in HP, we can express \(\frac{1}{\alpha}\), \(\frac{1}{\beta}\) and \(\frac{1}{\gamma}\) in terms of some constant value: \[\frac{1}{\alpha} = x - \frac{1}{\beta}\] \[\frac{1}{\gamma} = 2\frac{1}{\beta} - x\] Where x is a constant. Now we can express a, b, and c using these expressions: \[a = x - 1\] \[b = 2\] \[c = 3-x\]

Now we can substitute the values of a, b and c into the GP equation: \[(2\beta)^2=(x-1)\alpha(3-x)\gamma\]

Now we will simplify the equation and express it in terms of ratio. By simplifying, we get: \[\frac{1}{\alpha\gamma}=4\] Divide both sides by 4: \[\frac{1}{4}=\alpha\gamma\] Now using the previous expressions of \(\frac{1}{\alpha}\), \(\frac{1}{\beta}\) and \(\frac{1}{\gamma}\), we will express the ratio \(a:b:c\) in terms of their harmonic counterparts: \[\frac{x-1}{2}:\frac{2}{2}:\frac{3-x}{2}\] \[x-1:1:3-x\] Now, by comparing the coefficients with the given ratio, we get: \[\frac{1}{\gamma} = x-1\] \[\frac{1}{\beta} = 1 \] \[\frac{1}{\alpha} = 3-x\] These equations are consistent with the expressions we got in Step 2, confirming that the given ratio holds true: \(a:b:c = \frac{1}{\gamma}:\frac{1}{\beta}:\frac{1}{\alpha}\).