# Arithmetic and Geometric Progressions

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Type of Progression

## Type of Progression

If the equation px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root, then in which type of progression is $\frac{d}{p} , \frac{e}{q} , \frac{f}{r}$
1. $\frac{d}{p} , \frac{e}{q} , \frac{f}{r}$ are in G.P
2. $\frac{d}{p} , \frac{e}{q} , \frac{f}{r}$ are in A.P
3. $\frac{d}{p} , \frac{e}{q} , \frac{f}{r}$ are in H.P
4. Insufficient Data

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

## Detailed Solution

Since p,q,r are in G.P. we have q2 = pr. On solving px2 +2qx +r = 0 we get

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

=) x = $\frac{-2q}{2p}$ (q2 = pr)

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

Thus x = $\frac{-q}{p}$ is the repeated root of px2+2qx+r = 0

∴ x = $\frac{-q}{p}$ is also a root of dx2+2ex+f = 0

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

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

=) dq2 - 2eqp + fp2 = 0

=) $\frac{d}{p} - 2* \frac{e}{q} + \frac{fp}{q^2} = 0$ [On dividing by pq2]

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

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

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

Correct Answer: $\frac{d}{p} , \frac{e}{q} , \frac{f}{r}$ are in A.P

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## More questions from Progressions

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. Reinforce these ideas with these questions.