Approach:
Let the first term of the AP be a and the common difference be d.
Given, Sum of all the terms = 105
Also, greatest term = 6 times the least term
Therefore, the greatest term = a + (n-1)d = 6a
where n is the number of terms in the AP.

Calculation:
1. Sum of all terms in an AP:
Sum of n terms in an AP can be given by the formula:
Sum = (n/2)[2a + (n-1)d]
Here, Sum = 105
105 = (n/2)[2a + (n-1)d]

2. Greatest term is 6 times the least term:
a + (n-1)d = 6a
5a = (n-1)d

3. Substitute the value of d in equation 1:
105 = (n/2)[2a + 5a(n-1)/(n-1)]
105 = (n/2)[(7a-5a+5a(n-1))/(n-1)]
105 = (n/2)[(2a+5a(n-1))/(n-1)]

4. Simplify the equation:
210 = n[2a + 5a(n-1)]
Divide both sides by a:
210/a = n(2 + 5n - 5)
42/a = n(5n-3)

5. Check for values of a:
We need to find the lowest value of a.
From the above equation, we can see that a must be a factor of 42.
Therefore, the possible values of a are 1, 2, 3, 6, 7, 14, 21, 42.

6. Substitute values of a to find n:
For each value of a, we can find the corresponding value of n.
If the value of n is a positive integer, then that value of a is valid.
The values of a and n are as follows:
a = 1, n = 15
a = 2, n = 6
a = 3, n = 3.6 (not valid)
a = 6, n = 2.4 (not valid)
a = 7, n = 2.1 (not valid)
a = 14, n = 1.5 (not valid)
a = 21, n = 1.2 (not valid)
a = 42, n = 0.75 (not valid)

7. Find the common difference:
From the equation, 5a = (n-1)d, we get the common difference as:
d = 5a/(n-1)

8. Check for valid values of d:
If the value of d is positive, then the corresponding value of a is valid.
The values of a, n, and d are as follows:
a = 1, n = 15, d = 1/2 (valid)
a = 2, n = 6, d = 2/5 (not valid)
a = 3, n = 3.6, d = 3 (not valid)
a = 6, n = 2.4, d = 6 (not valid)
a = 7, n = 2.1, d