The question is about type of a progression. Roots of two quadratic progression is in a progression, what type it is? With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.

Question 19: If the equation px^{2} + 2qx + r = 0 and dx^{2} + 2ex + f = 0 have a common root, and p,q,r are in G.P., then in which type of progression is \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\)

- \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\) are in G.P
- \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\) are in A.P
- \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\) are in H.P
- Insufficient Data

\\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\) are in A.P

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Since p,q,r are in G.P. we have q^{2} = pr. On solving px^{2} +2qx +r = 0 we get

x = \\frac{-2q ± √(4q^2 - 4pr)}{2p}\\)

=> x = \\frac{-2q}{2p}\\) (q^{2} = pr)

=> x = \\frac{-q}{p}\\)

Thus x = \\frac{-q}{p}\\) is the repeated root of px^{2}+2qx+r = 0

∴ x = \\frac{-q}{p}\\) is also a root of dx^{2}+2ex+f = 0

=> d(\\frac{-q}{p}\\))^{2} + 2e(\\frac{-q}{p}\\)) + f = 0

=> \\frac{dq^2 - 2eqp + fp^2}{p^2}\\) = 0

=> dq^{2} - 2eqp + fp^{2} = 0

=> \\frac{d}{p}\\) - 2 * \\frac{e}{q}\\) + \\frac{fp}{q^2}\\) = 0 [On dividing by pq^{2}]

=> \\frac{d}{p}\\) - 2 * \\frac{e}{q}\\) + \\frac{f}{r}\\) = 0

=> \\frac{d}{p}\\) + \\frac{f}{r}\\) = 2 * \\frac{e}{q}\\)

Hence it can be clearly seen that \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\) are in A.P

The question is **"If the equation px ^{2} + 2qx + r = 0 and dx^{2} + 2ex + f = 0 have a common root, and p,q,r are in G.P., then in which type of progression is \\frac{d}{p}\\) , \\frac{e}{q}\\) , \\frac{f}{r}\\)"**

Choice B is the correct answer.

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