Question

A person takes a loan of Rs. 3,250. He paid Rs. 20 in the first month and then increased Rs. 15 in the payment of every month. Find in how many months he paid the loan.

Hint:

If payments form an arithmetic progression, use $S_n = \dfrac{n}{2}[2a + (n - 1)d]$ to find the total number of payments.

Answer 1

Step 1: Write the given data. 
Total loan amount $S_n = 3250$ 
First month payment $a = 20$ 
Common difference $d = 15$ 
Let the total number of months be $n$. 
Step 2: Use the formula for sum of $n$ terms of an arithmetic series. 
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \] Step 3: Substitute the known values. 
\[ 3250 = \frac{n}{2} [2(20) + (n - 1)(15)] \] \[ 3250 = \frac{n}{2} [40 + 15n - 15] \] \[ 3250 = \frac{n}{2} (15n + 25) \] \[ 6500 = n(15n + 25) \] 
Step 4: Simplify the equation. 
\[ 15n^2 + 25n - 6500 = 0 \] Divide by 5: \[ 3n^2 + 5n - 1300 = 0 \] 
Step 5: Solve the quadratic equation. 
Use the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, $a = 3$, $b = 5$, $c = -1300$. \[ n = \frac{-5 \pm \sqrt{5^2 - 4(3)(-1300)}}{2(3)} = \frac{-5 \pm \sqrt{25 + 15600}}{6} = \frac{-5 \pm \sqrt{15625}}{6} \] \[ \sqrt{15625} = 125 \Rightarrow n = \frac{-5 + 125}{6} = \frac{120}{6} = 20 \] Negative root is not possible. 
Step 6: Conclusion. 
\[ \boxed{n = 20} \] Hence, the loan was fully paid in 20 months.