Question

A recurring deposit account of $₹1200$ per month has a maturity value of $₹12440$. If the rate of interest is $8\%$ and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.

Answer 1

As per the given data

P$=$$₹$$1200$

Nmberofmonths$=$$n$

Interest rate$=$$8\%$

MaturityAmount$=$$₹$$12440$

Find out

We have to determine the time of this Recurring Deposit Account.

Solution

MaturityValue$\mathbf{=}\mathit{P}\mathbf{\times }\mathit{n}\mathbf{+}\mathit{P}\mathbf{\times }\left[n\left(\frac{n+1}{2\times 12}\right)\right]\mathbf{\times }\frac{\mathbf{r}}{\mathbf{100}}$

On substituting the known values in the above equation we get

$$\begin{gathered} 12440=1200\times n+1200\times \left[n\left(\frac{n+1}{2\times 12}\right)\right]\times \frac{8}{100} \\ \Rightarrow 12440=1200n+1200\times n\left(\frac{n+1}{24}\right)\times \frac{8}{100} \\ \Rightarrow 12440=1200n+50n\left(n+1\right)\times \frac{4}{50} \\ \Rightarrow 12440=1200n+4n\left(n+1\right) \\ \Rightarrow n^2+301n-3110=0 \end{gathered}$$

On factorising

$$\begin{gathered} \Rightarrow n^2+\left(311-10\right)n-3110=0 \\ \Rightarrow n^2+311n-10n-3110=0 \\ \Rightarrow n\left(n+311\right)-10\left(n+311\right)=0 \\ \Rightarrow \left(n+311\right)\left(n-10\right)=0 \\ \Rightarrow n=-311orn=10 \end{gathered}$$

n cannot be negative we get

$n=10$

Therefore, $10$ months are required