Given: Focal length f = 40 cm (converging lens, so positive), object distance u = 160 cm.

Given: Focal length $f = 40 \text{ cm}$ (converging lens, so positive), object distance $u = 160 \text{ cm}$.
The lens formula is:

Substitute the given values:

Rearrange to solve for $\frac{1}{v}$:

Find a common denominator:

Subtract the fractions:

Invert to find $v$:

The position of the image is $\boxed{\text{53.33 cm from the lens on the opposite side of the object}}$.

Given: Object height $h_o = 120 \text{ cm}$.
The magnification formula is:

Substitute the values for $v$ and $u$:

Now, calculate the image height $h_i$:

The size of the image is $\boxed{\text{40 cm}}$.

Since $v$ is positive, the image is formed on the opposite side of the lens from the object, indicating a real image.
Since $M$ is negative, the image is inverted.
Since $|M| = \frac{1}{3} < 1$, the image is diminished.
The nature of the image is $\boxed{\text{real, inverted, and diminished}}$.