A manufacturer reckons that the value of a machine which costs him ₹ 31,250 depreciates each year by 20%. Find its value after 2 years.

Given,

The initial cost of machine is ₹ 31,250. Therefore,

a = ₹ 31,250

Depreciation Rate : 20% per year

If 20% is lost, the percentage of the value retained is: 100% - 20% = 80%.

r = $\dfrac{80}{100}$ [constant factor by which the value of the machine is multiplied each year to get the next year's value.]

Since, value after 2 years is the value in the beginning of third year, thus n = 3.

We know that,

$\Rightarrow T$n = ar^{(n-1)} \\[1em] \Rightarrow T3 = ar^{3-1} \\[1em] = 31250 \Big(\dfrac{80}{100}\Big)^2 \\[1em] = 31250 (0.8)^2 \\[1em] = 31250 (0.64) \\[1em] = 20,000.

Hence, value of machine after 2 years = ₹20,000.