A customer pays ₹975 in instalments. The payment is done each month ₹5 less than the previous month. If the first
instalment is ₹100, how much time will be taken to pay the entire amount?

Question

Accepted Answer

Given: First instalment a = ₹100, each subsequent instalment decreases by ₹5, total amount paid = ₹975.

This forms an arithmetic progression (AP) with first term a = 100 and common difference d = -5.

Let the number of instalments be n.

Sum of n terms of AP, S_n = n/2 * [2a + (n-1)d]

Substitute the values:  
975 = n/2 * [2*100 + (n-1)*(-5)]  
975 = n/2 * [200 - 5(n-1)]  
975 = n/2 * [200 - 5n + 5]    
975 = n/2 * (205 - 5n)

Multiply both sides by 2:   
1950 = n(205 - 5n)

Expand:    
1950 = 205n - 5n^2

Rearranged:   
5n^2 - 205n + 1950 = 0

Divide entire equation by 5:   
 n^2 - 41n + 390 = 0

Solve quadratic equation:  
Discriminant D = 41^2 - 4*1*390 = 1681 - 1560 = 121

Roots:    
n = [41 ± sqrt(121)] / 2 = (41 ± 11)/2

Two possible values:   
n = (41 + 11)/2 = 52/2 = 26   
n = (41 - 11)/2 = 30/2 = 15

Check which value fits the problem:    
If n=15, sum = 15/2 * [2*100 + 14*(-5)] = 7.5 * (200 - 70) = 7.5 * 130 = 975

If n=26, sum = 26/2 * [200 + 25*(-5)] = 13 * (200 - 125) = 13 * 75 = 975

Both 15 and 26 months sum to 975, but the instalments decrease by ₹5 each month starting from ₹100. After 15 months, the instalment amount is:    
100 - (15-1)*5 = 100 - 70 = ₹30

After 26 months, the instalment would be:    
100 - (26-1)*5 = 100 - 125 = -₹25 (not possible)

Hence, the valid number of months is 15.

Therefore, the customer will take 15 months to pay the entire amount.

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