Let the initial number of people as 'x'
The total initial weight as 'W'
Initial average weight = 55 kg
New person's weight = 65 kg
Increase in average weight = 2 kg
\(\frac{W}{x} = 55\) (Initial average weight) —>eq1
\(\frac{W + 65}{x + 1} = 57\) (New average weight)---> eq2
From eq 1, W = 55x
Substituting W in eq2:
\(\frac{55x + 65}{x + 1} = 57\)
=> 55x + 65 = 57x + 57
=> 2x = 8
=> x = 4
So, the initial number of people (x) is 4.
Total number of people in the new group = x + 1 = 4 + 1 = 5
Therefore, the ratio of the number of people initially in the group to the total number of people in the new group is 4:5.
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: