Given information:
- The sum of the first 6 terms of an arithmetic progression (AP) is 42.
- The ratio of the 10th term to the 30th term is 3:1.

Let's solve the problem step by step:

Step 1: Find the common difference (d) of the AP.
To find the common difference, we need to use the formula for the sum of an AP:
Sn = (n/2)(2a + (n-1)d)
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.

Given that the sum of the first 6 terms is 42, we can substitute the values into the formula:
42 = (6/2)(2a + (6-1)d)
42 = 3(2a + 5d)
14 = 2a + 5d

Step 2: Find the values of a and d using the ratio of terms.
The ratio of the 10th term to the 30th term is 3:1. We can write this as an equation:
(10th term) / (30th term) = 3/1

Using the formula for the nth term of an AP, we can write the equation as:
(a + 9d) / (a + 29d) = 3/1

Cross-multiplying and simplifying the equation, we get:
a + 9d = 3(a + 29d)
a + 9d = 3a + 87d
2a = -78d
a = -39d

Step 3: Substitute the value of a into the equation from Step 1 to find d.
We already know that a = -39d. Substituting this into the equation 14 = 2a + 5d, we get:
14 = 2(-39d) + 5d
14 = -78d + 5d
14 = -73d
d = -14/73

Step 4: Substitute the value of d into the equation a = -39d to find a.
Using the value of d = -14/73, we can find a:
a = -39(-14/73)
a = 6

Step 5: Find the 13th term of the AP.
Using the formula for the nth term of an AP, we can find the 13th term:
13th term = a + (13-1)d
13th term = 6 + 12(-14/73)
13th term = 6 - 168/73
13th term = 330/73

Step 6: Summarize the results.
The first term of the AP is 6, and the 13th term is 330/73.

Nice question