Question

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly instalments. Each instalment being less than the preceding one by Rs 40. Find the amount of the first and last instalment.

Answer 1

Money he borrows = Rs 8000
Total interest = Rs 1360

Total money he will pay after 12 months = Rs (8000 + 1360) = Rs 9360

It is given that,
Each instalment being less than the preceding one by Rs 40.

d = −40
n = 12
S₁₂ = 9360

Now,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ {S}_{12}=\frac{12}{2}\left(2a+\left(12-1\right)d\right) \\ \Rightarrow 9360=\frac{12}{2}\left(2\left(a\right)+\left(12-1\right)\left(-40\right)\right) \\ \Rightarrow 9360=6\left(2a-440\right) \\ \Rightarrow 2a-440=\frac{9360}{6} \\ \Rightarrow 2a-440=1560 \\ \Rightarrow 2a=1560+440 \\ \Rightarrow 2a=2000 \\ \Rightarrow a=1000 \end{gathered}$$

Also,
$$\begin{gathered} a_{12}=a+\left(12-1\right)d \\ =1000+11\left(-40\right) \\ =1000-440 \\ =560 \end{gathered}$$

Hence, the first and last instalment are Rs 1000 and Rs 560, respectively.