Given:
- The ratio of the first term (Rst) to the last term (Lst) of an arithmetic progression is 3:4.
- The sum of all terms in the arithmetic progression is 210.
- There are a total of 6 terms in the arithmetic progression.

To find:
- The common difference (d) of the arithmetic progression.

Approach:
1. Let's assume that the first term of the arithmetic progression is a and the common difference is d.
2. Therefore, the last term (Lst) can be represented as a + (n-1)d, where n is the total number of terms in the arithmetic progression.
3. Given that the ratio of Rst to Lst is 3:4, we can write the equation as:
a/Lst = 3/4
Cross-multiplying, we get 4a = 3Lst
Dividing both sides by 3, we get a = (3/4)Lst

Deriving the Formula for Sum of an Arithmetic Progression:
4. The sum of an arithmetic progression can be calculated using the formula:
S = (n/2)(a + Lst), where S is the sum, n is the number of terms, a is the first term, and Lst is the last term.
5. Substituting the values, we have:
210 = (6/2)(a + a + 5d)
Simplifying, we get:
210 = 3(2a + 5d)
Dividing both sides by 3, we get:
70 = 2a + 5d

Solving the Equations:
6. We have two equations:
a = (3/4)Lst --(1)
70 = 2a + 5d --(2)
7. Substituting the value of a from equation (1) into equation (2), we get:
70 = 2((3/4)Lst) + 5d
Simplifying, we get:
70 = (3/2)Lst + 5d
Multiplying both sides by 2, we get:
140 = 3Lst + 10d
8. We also know from the given information that the ratio of Rst to Lst is 3:4. Therefore, we can write:
Rst/Lst = 3/4
Cross-multiplying, we get:
4Rst = 3Lst
Dividing both sides by 3, we get:
Rst = (3/4)Lst
Substituting the value of Rst from this equation into equation (8), we get:
140 = 3(4Rst) + 10d
Simplifying, we get:
140 = 12Rst + 10d
9. Since we have two equations with two variables, we can solve them simultaneously to find the values of Rst and d.

Solving the Simultaneous Equations:
10. We have the equations:
140 = 12Rst + 10d --(3)
70