Find the relation between x and y in order that the $r^{th}$ mean between x and 2y may be the same as the $r^{th}$ mean between $2 x$ and $y, n$ means being inserted in each case.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

y = (\frac{n - r + 1}{r}) x

y = (\frac{n - r + 2}{r}) x

y = (\frac{n - r + 3}{r}) x

y = (\frac{n - r + 4}{r}) x

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In case of n means inserted between two terms a and b, then the common difference of will be,

d = \frac{b - a}{n + 1}

Here, the value of

r^{th} mean,

= a + rd

Here, we have two terms x and y. So,

x + r \times (\frac{2 y - x}{n + 1}) = 2 x + r \times (\frac{y - 2 x}{n + 1})

\Rightarrow (n + 1) x + 2 ry - rx = 2 (n + 1) x + ry - 2 rx

\Rightarrow y = (\frac{n - r + 1}{r}) x

This is the required relation.
Thus, the correct option is 1.

Watch 3-min video & get full concept clarity