Question 1162673
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            The problem's formulation is crystally clear. 


            See my solution below.



<pre>
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

    {{{38.5/88}}} = {{{77/176}}} = {{{7/16}}}.    <U>ANSWER</U>
</pre>

Solved.