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The problem's formulation is crystally clear.
See my solution below.
Let a is the first term and d is the common difference of the first AP.
Let b is the first term and e is the common difference of the second AP.
Then
( (n/2)*(2a+(n−1)*d) ) / ( (n/2)*(2b+(n−1)*e) ) = (3n+8)/(7n+15)
or
(2a+(n−1)*d)/(2b+(n−1)*e) = (3n+8)/(7n+15)
or
((2a-d) + nd)/((2b-e)+ed) = (3n+8)/(7n+15).
It means that up to a coefficient of proportionality
2a−d = 8 and d = 3, which implies 2a = 11, d =3, or a = 5.5, d = 3
also
2b−e = 15 and e = 7, which implies 2b = 22, e = 7, or b = 11, e = 7.
Now the 12-th terms of the progressions are (up to proportionality)
5.5 + 11*3 = 38.5 and 11 + 7*11 = 88
and their ratio is
= = . ANSWER
Solved.