Given:
- Sum of the first q terms of an AP is 162.
- The ratio of the 6th term and the 13th term is 1:2.

To Find:
- First term (a) and 15th term (l) of the AP.

Explanation:

1. Sum of the first q terms of an AP:
The sum of the first q terms of an arithmetic progression (AP) is given by the formula:
S_q = (q/2)(2a + (q-1)d)

In this case, the sum of the first q terms is given as 162. So, we have:
162 = (q/2)(2a + (q-1)d) ...........(1)

2. Ratio of the 6th term and 13th term:
The formula to find the nth term (T_n) of an AP is:
T_n = a + (n-1)d

Given that the ratio of the 6th term (T₆) and the 13th term (T₁₃) is 1:2, we can write:
T₆/T₁₃ = 1/2

Using the formula for the nth term, we can substitute the values:
(a + 5d)/(a + 12d) = 1/2 ...........(2)

3. Solving the Equations:
To find the first term (a) and the common difference (d), we need to solve equations (1) and (2) simultaneously.

From equation (2), we can rewrite it as:
2(a + 5d) = a + 12d
2a + 10d = a + 12d
a = 2d ...........(3)

Substituting the value of a from equation (3) into equation (1), we get:
162 = (q/2)(4d + (q-1)d)
324 = q(4d + (q-1)d)
324 = q(5d + (q-1)d)
324 = q(5d + qd - d)
324 = q(6qd - d)
324 = q(6qd - d)
324 = 6qd^2 - qd
6qd^2 - qd - 324 = 0

Now, we can solve this quadratic equation to find the values of q and d. Once we have q and d, we can substitute them into equation (3) to find the first term (a).

4. Finding the 15th term:
Once we have the values of a and d, we can use the formula for the nth term to find the 15th term (T₁₅):
T₁₅ = a + (15-1)d

5. Conclusion:
By solving the equations and substituting the values, we can find the first term (a), the 15th term (T₁₅), and the common difference (d) of the arithmetic progression (AP).