The ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is 6/13. If each term of this sequence is positive, and the product of the first term and the third term is 32, find the sum of the first 100 terms of this sequence.

In an arithmetic progression (AP), the nth term can be given by the formula a + (n-1)d, where a is the first term and d is the common difference.

Given that the ratio of the fifth term to the twelfth term is 6/13, we can write this as:

(a + 4d) / (a + 11d) = 6/13

Cross multiplying gives us:

13a + 52d = 6a + 66d

Solving this equation for a gives us:

a = 14d

We are also given that the product of the first term and the third term is 32, so we can write this as:

a * (a + 2d) = 32

Substituting a = 14d into this equation gives us:

14d * (14d + 2d) = 32

Solving this equation for d gives us:

d = 1/2

Substituting d = 1/2 into a = 14d gives us:

a = 7

The sum of the first n terms of an AP can be given by the formula n/2 * (2a + (n-1)d). Substituting a = 7, d = 1/2, and n = 100 into this formula gives us:

S = 100/2 * (2*7 + (100-1)*1/2) = 50 * (14 + 99/2) = 50 * (14 + 49.5) = 50 * 63.5 = 3175

So, the sum of the first 100 terms of this sequence is 3175.