Question:
A man repays a loan of Rs. 3250 by paying Rs. 20 in the first month and then increases the payment by Rs. 15 every month. The number of months it takes to clear the loan is
A man repays a loan of Rs. 3250 by paying Rs. 20 in the first month and then increases the payment by Rs. 15 every month. The number of months it takes to clear the loan is
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Loan repayment problems often reduce to AP sum equations.
Updated On: Apr 30, 2026
- $20$
- $25$
- $35$
- $40$
- $10$
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept: Arithmetic Progression Sum
Payments form AP:
\[
a=20, d=15
\]
Step 1: Sum formula \[ S_n = \frac{n}{2}[2a + (n-1)d] \]
Step 2: Substitute \[ 3250 = \frac{n}{2}[40 + 15(n-1)] \] \[ = \frac{n}{2}[40 + 15n - 15] = \frac{n}{2}(25 + 15n) \]
Step 3: Multiply \[ 6500 = n(25 + 15n) \] \[ 6500 = 25n + 15n^2 \] \[ 15n^2 + 25n - 6500 = 0 \]
Step 4: Divide by 5 \[ 3n^2 + 5n -1300 =0 \]
Step 5: Solve quadratic \[ n = \frac{-5 \pm \sqrt{25 + 15600}}{6} \] \[ = \frac{-5 \pm 125}{6} \] \[ n = 20 (\text{positive root}) \] \[ \boxed{20} \]
Step 1: Sum formula \[ S_n = \frac{n}{2}[2a + (n-1)d] \]
Step 2: Substitute \[ 3250 = \frac{n}{2}[40 + 15(n-1)] \] \[ = \frac{n}{2}[40 + 15n - 15] = \frac{n}{2}(25 + 15n) \]
Step 3: Multiply \[ 6500 = n(25 + 15n) \] \[ 6500 = 25n + 15n^2 \] \[ 15n^2 + 25n - 6500 = 0 \]
Step 4: Divide by 5 \[ 3n^2 + 5n -1300 =0 \]
Step 5: Solve quadratic \[ n = \frac{-5 \pm \sqrt{25 + 15600}}{6} \] \[ = \frac{-5 \pm 125}{6} \] \[ n = 20 (\text{positive root}) \] \[ \boxed{20} \]
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