The given question is related to Arithmetic Progressions (AP). We are given that the ratio of the sum of n terms of two APs is (n 1) : (n-1). We need to find the ratio of their mth term.

Let's assume the first AP has a common difference of 'd1' and the second AP has a common difference of 'd2'.

To find the sum of n terms of an AP, we use the formula Sn = (n/2) * (2a + (n-1)d), where 'a' is the first term and 'd' is the common difference.

Therefore, the sum of n terms for the first AP is given by S1 = (n/2) * (2a1 + (n-1)d1), and for the second AP, it is given by S2 = (n/2) * (2a2 + (n-1)d2).

Now, we are given that the ratio of S1 to S2 is (n 1) : (n-1). So, we can write the equation:

(n/2) * (2a1 + (n-1)d1) / (n/2) * (2a2 + (n-1)d2) = (n 1) / (n-1)

Simplifying this equation, we get:

(2a1 + (n-1)d1) / (2a2 + (n-1)d2) = (n 1) / (n-1)

Now, let's find the ratio of the mth term of the two APs.

The mth term of an AP is given by Tm = a + (m-1)d.

For the first AP, the mth term is T1 = a1 + (m-1)d1, and for the second AP, it is T2 = a2 + (m-1)d2.

To find the ratio of T1 to T2, we can write the equation as:

(a1 + (m-1)d1) / (a2 + (m-1)d2)

Now, let's substitute the value of a1 and a2 from the equations for Sn:

(a1 + (m-1)d1) = (2a1 + (n-1)d1) / (n-1) * (a2 + (m-1)d2)

Simplifying this equation, we get:

(a1 + (m-1)d1) = (2a2 + (n-1)d2) / (n 1) * (a2 + (m-1)d2)

Thus, the ratio of the mth term of the two APs is (2a2 + (n-1)d2) / (n 1) * (a2 + (m-1)d2).

In conclusion, the ratio of the mth term of the two APs is (2a2 + (n-1)d2) / (n 1) * (a2 + (m-1)d2).