Question

A man repays a loan of $₹3250$ by paying $₹20$ in the first month and then increase the payment by $₹15$ every month. How long will it take him to clear the loan?

Answer 1

Step1: Framing an equation for $n$.

The amount of the loan paid is in the form of an A.P. series, having first term as $a=20$ and common difference as $d=15$.

Let the loan is paid in $n$ number of instalments.

Then, the total loan to be paid i.e.,$₹3250$ is the sum of $n$ number of terms of the A.P.

The sum of $n$ number of terms of an A.P. is given by $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$, where $a$ is the first term and $d$ is a common difference.

Substitute the values of $S_n,a$ and $d$ in the formula $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$ to calculate the value of $n$.

$$\begin{gathered} \Rightarrow 3250=\frac{n}{2}\left[2\left(20\right)+(n-1)15\right](Substitutingthevaluesofa,dand{S}_n) \\ \Rightarrow 6500=n\left[40+15n-15\right](Multiplyingbothsidesby2) \\ \Rightarrow 6500=n\left[25+15n\right] \\ \Rightarrow 6500=25n+15n^2 \\ \Rightarrow 1300=5n+3n^2(Dividingbothsidesby5) \\ \Rightarrow 3n^2+5n-1300=0 \\ \therefore (n-20)(3n+65)=0(bysplittingmiddletermandfactori\sin g) \end{gathered}$$

Step2: Calculation of the value of $n$.

Equate both the factors to 0 to obtain the solution.

$$\begin{gathered} n-20=0 \\ \Rightarrow n=20 \\ 3n+65=0 \\ \Rightarrow 3n=-65(subtracting65frombothsides) \\ \therefore n=-\frac{65}{3}(dividingbothsidesby3) \end{gathered}$$

Since, the number of terms can not be negative, therefore, the correct value of $n$ is 20.

Final Answer: The loan will be paid in 20 months.