To find the value of mn, we start with the general term in the expansion of (mx+‌
1
x
)n, given by the binomial theorem: Tk+1=(
n
k
)(mx)n−k(‌
1
x
)k We need to find the 4 th term in this expansion, so let k=3. Substituting k=3 into the general term formula, we get: T4=(
n
3
)(mx)n−3(‌
1
x
)3 Simplify the expression: T4=(
n
3
)mn−3xn−3⋅‌
1
x3
Combine the exponents of x : T4=(
n
3
)mn−3xn−3−3=(
n
3
)mn−3xn−6 Given that the 4 th term is ‌
5
2
, this implies: (
n
3
)mn−3xn−6=‌
5
2
For the term to be a constant (i.e., not dependent on x ), the exponent of x must be zero: n−6=0⇒n=6 Substitute n=6 into the equation for the 4th term: T4=(
6
3
)m6−3x0=‌
5
2
This simplifies to: (
6
3
)m3=‌
5
2
Calculate (
6
3
) : (
6
3
)=‌
6!
3!3!
=20 Substitute (
6
3
) into the equation: 20m3=‌
5
2
Solve for m3 : m3=‌
5
2
×‌
1
20
=‌
5
40
=‌
1
8
Taking the cube root of both sides, we find: m=‌