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

Total amount to be paid, $$S_n$$​ = ₹975

The first installment, a = ₹100

Each subsequent installment(d) is ₹5 less than the previous month.

The sum of the first n terms of an arithmetic progression (AP) is given by:

​$$S_n = \frac{n}{2} \times [2a + (n-1) \times d]$$​​

Where:

$$S_n$$ is the total sum,

n is the number of installments,

a is the first installment,

d is the common difference.

Substituting values into the sum formula:

​$$975 = \frac{n}{2} \times [2 \times 100 + (n - 1) \times (-5)] \\975 = \frac{n}{2} \times [200 - 5(n - 1)] \\975 = \frac{n}{2} \times [200 - 5n + 5] \\975 = \frac{n}{2} \times [205 - 5n] \\1950 = n \times (205 - 5n) \\1950 = 205n - 5n^2 \\5n^2 - 205n + 1950 = 0 \\n^2 - 41n + 390 = 0 \\n^2 – 26n -15n + 390 = 0 \\n(n – 26 ) – 15( n – 26) = 0 \\(n – 26)(n – 15) = 0$$​​

n = 26 or n = 15

n = 15 and n = 26.  

As n has two values, 15 and 26 but only 15 is correct in this condition because After 15 month he can't pay less than ₹0 in last installment. After Zero all the numbers in the series became negative which cannot be installment.

Thus, option(b) 15 month is correct.