Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Question

If the mean deviation about the median of the numbers \[ k,\,2k,\,3k,\,\ldots,\,1000k \] is \(500\), then \(k^2\) is equal to

Answer 1

We are given that two numbers are to be inserted between 3 and 81 such that the resulting sequence is a geometric progression.

Let the required G.P. be \[ 3,\; x,\; y,\; 81 \]

Let the common ratio be r.

Then, \[ 3r^3 = 81 \]

\[ r^3 = 27 \]

\[ r = 3 \]

Hence, the G.P. becomes \[ 3,\; 3 \times 3,\; 3 \times 3^2,\; 81 \]

\[ 3,\; 9,\; 27,\; 81 \]

Therefore, the two numbers to be inserted are \[ 9 \text{ and } 27. \]

Solution and Explanation