Introduction:
The given sequence s is defined by the formula s_n = 1 * (3/n), where n = 1, 2, 3, ...

Finding the terms:
To find the terms of the sequence, we substitute the values of n into the formula. Let's calculate the first few terms:

s_1 = 1 * (3/1) = 3
s_2 = 1 * (3/2) = 1.5
s_3 = 1 * (3/3) = 1

Product of consecutive terms:
To find the product of consecutive terms from the 4th term to the 60th term, we need to calculate the product of s_4 * s_5 * s_6 * ... * s_60.

Let's calculate the product of the first few terms as an example:

s_4 = 1 * (3/4) = 0.75
s_5 = 1 * (3/5) = 0.6
s_6 = 1 * (3/6) = 0.5

Product of s_4 * s_5 * s_6 = 0.75 * 0.6 * 0.5 = 0.225

General formula for the nth term:
We can observe that the nth term of the sequence s can be written as s_n = 1 * (3/n) = 3/n.

Product of consecutive terms:
To find the product of consecutive terms from the 4th term to the 60th term, we can write it as:

s_4 * s_5 * s_6 * ... * s_60 = (3/4) * (3/5) * (3/6) * ... * (3/60)

We can cancel out the common factors of 3 in the numerator and denominator:

(3/4) * (3/5) * (3/6) * ... * (3/60) = 1/4 * 1/5 * 1/6 * ... * 1/20

Cancelling out common factors:
Now, let's write the product in a simplified form by cancelling out the common factors:

1/4 * 1/5 * 1/6 * ... * 1/20 = (1/4 * 1/5) * (1/6 * 1/7) * ... * (1/19 * 1/20)

We can see that each fraction in the product can be written as a pair of adjacent fractions:

(1/4 * 1/5) * (1/6 * 1/7) * ... * (1/19 * 1/20) = [(1/4 * 1/5) * (1/6 * 1/7)] * ... * [(1/18 * 1/19) * (1/20)]

We can observe that each pair of adjacent fractions cancels out:

(1/4 * 1/5) * (1/6 * 1/7) * ... * (1/19 * 1/20) = 1/20

Conclusion:
The product of