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: Here we make it as a series of $AP$ because we add 15 in every month. Let 15 be a common difference. Then apply the formula of summation of n terms of i.e. ${S_n} = \dfrac{n}{2}(a + l){\text{ }}$

Complete step-by-step answer:
First month he paid$\; = 20Rs$.
Second month he pays ${\text{ = 20 + 15}}$
Third month he pays $= \;20 + 2 \times 15$ Here we see every month Rs. 15 is increase so we can say it is forming an AP whose nth term can be written as ${T_n} = a + (n - 1)d$ (where a=20 and d=15)
So, $n'th$ Month he pays $= \;20 + (n - 1) \times 15$
According to the question he increases his payment by $Rs.15$ every month
Now, total payment paid by him $= \;(20) + (20 + 15) + (20 + 2 \times 15) + ... + 20 + (n - 1) \times 15 = 3250$
This in the form of summation of $n$ term of$AP$.
So we apply the formula of summation of $n$ term of$AP$.
${S_n} = \dfrac{n}{2}(a + l){\text{ }}$ Here $a =$first term, $l =$last term.
$\Rightarrow \dfrac{n}{2}\left( {(20) + 20 + (n - 1) \times 15} \right) = 3250 \\\ \Rightarrow 40n + 15{n^2} - 15n = 6500 \\\ \Rightarrow 15{n^2} + 25n - 6500 = 0 \\\ \Rightarrow 3{n^2} + 5n - 1300 = 0 \\\ \Rightarrow (n - 20)(3n + 65) = 0 \\\$
$\Rightarrow n = 20, - \dfrac{{65}}{3}$ Here $- \dfrac{{65}}{3}{\text{ }}$ is not possible because day cannot be negative.
Hence, $n{\text{ = 20}}$ months answer.

Note: Whenever we face such a type of question the key concept for solving this question is make the series as the question says then decide the series. Then apply the formula of that series to get the answer.