Problem Statement:
Divide 156 into 4 parts such that they are in continued proportion and the sum of the first and third part is in a ratio of 1:5 to the sum of the second and fourth parts.

Solution:
Let's assume that the four parts are a, ar, ar^2 and ar^3, where a is the first part and r is the common ratio.

Step 1:
We know that the sum of the first and third part is in a ratio of 1:5 to the sum of the second and fourth parts. So, we can write the equation as:
a + ar^2 = k(ar + ar^3), where k = 1/5

Step 2:
We also know that the four parts are in continued proportion, so we can write another equation as:
a/ar = ar/ar^2 = ar^2/ar^3

Simplifying this equation, we get:
a = ar^3
ar = ar^2
ar^2 = a

Step 3:
Using the above equations, we can find the value of r:
ar^2 = a
r^2 = 1
r = ±1

Step 4:
If r = 1, then a = ar^3 = a, which means all the four parts are equal. But this doesn't satisfy the condition that the sum of the first and third part is in a ratio of 1:5 to the sum of the second and fourth parts.

Step 5:
Therefore, r cannot be equal to 1. So, let's assume that r = -1. Using this value of r, we can find the values of a and the other three parts:

a + ar + ar^2 + ar^3 = 156
a(1-1+1-1) = 156
a = 39

Therefore, the four parts are:
a = 39
ar = -39
ar^2 = 39
ar^3 = -39

Step 6:
We can check whether the sum of the first and third part is in a ratio of 1:5 to the sum of the second and fourth parts:

a + ar^2 = 39 + 39 = 78
ar + ar^3 = -39 - 39 = -78

(78: -78) = (1:5)

Therefore, the four parts are 39, -39, 39, and -39, and they satisfy the given conditions.