Question

A sum of Rs 1000 is invested at 8% per annum simple interest. Calculate the interest at the end of each year. Show that these interest form an AP. Using this fact, find the interest at the end of 30 years.

Answer 1

$$\begin{gathered} \mathrm{Simple}\mathrm{interest}=\frac{\left(P\times R\times T\right)}{100},\mathrm{where}\mathrm{P}=\mathrm{principal},\mathrm{R}=\mathrm{rate}\mathrm{of}\mathrm{interest}\mathrm{and}\mathrm{T}=\mathrm{time}\mathrm{in}\mathrm{years}. \\ =\frac{\left(1000\times 8\times \mathrm{T}\right)}{100}=80\times \mathrm{T} \end{gathered}$$
On putting T = 1, 2, 3, 4, we get:
Rs 80, Rs 160, Rs 240, Rs 320 and so on.

The difference between the interests of two consecutive years = (160 - 80) = (240 - 160) = Rs 80 [Constant]
So, the progression of simple interests forms an AP.
Here, first term = Rs 80
Common difference = Rs 80

Now, interest at the end of 30 years, T₃₀ = a + (n - 1)d
= 80 + (30 - 1)(80) = 80 + 29(80) = Rs 2400
Hence, the interest at the end of 30 years is Rs 2400.