To solve this problem, let's consider the given information:
- The lengths of two larger sides of the triangle are 10 and 9.
- The angles of the triangle are in an arithmetic progression (AP).

Let's assume that the three angles of the triangle are A, B, and C, with A being the smallest angle. Since the angles are in AP, we can represent them as A-d, A, and A+d, where d is the common difference.

Now, let's use the law of cosines to find the length of the third side of the triangle. According to the law of cosines, the square of the length of the third side (c^2) is equal to the sum of the squares of the other two sides (a^2 + b^2) minus twice their product (2ab) multiplied by the cosine of the angle opposite the third side (C).

c^2 = a^2 + b^2 - 2ab*cos(C)

Substituting the given values, we have:

c^2 = 9^2 + 10^2 - 2*9*10*cos(A+d)

Now, since the angles are in AP, we can rewrite cos(A+d) as cos(A) using the formula for the sum of angles in AP:

cos(A+d) = cos(A)cos(d) - sin(A)sin(d)

Substituting this into the equation, we have:

c^2 = 9^2 + 10^2 - 2*9*10(cos(A)cos(d) - sin(A)sin(d))

Simplifying further, we get:

c^2 = 9^2 + 10^2 - 18*cos(A)*cos(d) + 18*sin(A)*sin(d)

Now, let's analyze the options given and find the one that satisfies the equation:

a) 5
b) 7
c) 5√3
d) 5√6

We can see that option D, 5√6, is the only one that contains both a cosine and a sine term. Therefore, it is the correct answer.

In summary, by using the law of cosines and the fact that the angles are in an arithmetic progression, we can determine that the length of the third side of the triangle is 5√6.

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