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. How long will it take him to clear the loan?

Answer 1

Hint: The given question requires application of formula of summation of arithmetic progression (AP). Since there is a constant increase of payment, it forms an arithmetic sequence in the given case. We will find the number of months with the summation formula of \[n\] terms where the total amount will be equal to \[3250\].

Complete step-by-step answer:
Formula to find the sum up to \[{n^{th}}\]term of AP is given as:
\[{S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\} \]
Here \[a\]is the first term, \[d\] is the common difference and \[n\] is the number of terms.
Let us form an AP with the information given in the question:
The payment is Rs. \[20\]in first month, which increases by Rs. \[15\] every month. Hence, we get,
\[a = 20\]and \[d = 15\].
AP will be as follows:
\[20,35,50,65,....\]
We are given that total amount of repayment is Rs. \[3250\]. Hence, we can conclude that
\[{S_n} = 3250\]
Substituting the information in the formula, we get,
\[3250 = \dfrac{n}{2}\{ 2(20) + (n - 1)(15)\} \]
Solving the brackets, we get,
\[3250 = \dfrac{n}{2}\{ 40 + 15n - 15\} \]
\[3250 = \dfrac{n}{2}(25 + 15n)\]
Multiplying with \[2\] on the left-hand side of equation, we get,
\[6500 = n(25 + 15n)\]
Forming quadratic equation, we get,
\[15{n^2} + 25n - 6500 = 0\]

Dividing the equation by \[5\], we get,
\[3{n^2} + 5n - 1300 = 0\]
Factorising the equation, we get,
\[3{n^2} - 60n + 65n - 1300 = 0\]
\[3n(n - 20) + 65(n - 20) = 0\]
\[(n - 20)(3n + 65) = 0\]
Hence \[n = 20\]or \[n = - \dfrac{{65}}{3}\]
But since months cannot be negative, we get
\[n = 20\]
Therefore, the man will take \[20\]months to clear the loan.

Note: Arithmetic sequence is one where the difference between any two consecutive numbers is constant. In the given question, we can also find the amount the man in last month using the formula for \[{n^{th}}\]term of an AP:
\[{t_n} = a + (n - 1)d\]
For example, in the given sum, the last amount paid will be:
\[{t_{20}} = 20 + (20 - 1)(15) = 20 + 19(15) = 305\]