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? \[\]
A.40 months \[\]
B. 60 months \[\]
C. 20 months \[\]
D. 30 months \[\]

Answer 1

Hint: We assume the the man clears the loan in $ {{n}^{\text{th}}} $ month. We find the total amount paid up to $ {{n}^{\text{th}}} $ month as using sum up to $ n $ terms of an AP $ {{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\} $ where $ a $ the amount is is paid in first month and $ d $ is the increase amount every month and equate to given 3250. We solve for $ n $ to get the required number of months. \[\]

Complete step by step answer:
We know that the sum up to $ {{n}^{\text{th}}} $ of an arithmetic progression with first term $ a $ and common difference $ d $ is given by
\[{{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}\]
We are given in the question that the 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. So in second month he pays $ 20+15=35 $ rupees and in the third month he pays $ 35+15=50 $ rupees and so on. So we get sequence of payments as
\[20,35,50,...\]
We see that the above sequence is an AP sequence with common difference $ d=35-20=50-35=15 $ and first term $ a=20 $ . Let us assume that he clears the loan in $ {{n}^{\text{th}}} $ month which means he pays back complete 3250 rupees. So the sum of the amounts from the first month to $ {{n}^{\text{th}}} $ month will be equal to 3250 rupees. We use the sum up to $ {{n}^{\text{th}}} $ term formula for an AP and equate the sum to 3250. We have;
\[\begin{align}
  & 3250=\dfrac{n}{2}\left\{ 2\times 20+\left( n-1 \right)15 \right\} \\
 & \Rightarrow 6500=n\left( 15n+25 \right) \\
 & \Rightarrow 15{{n}^{2}}+25n-6500=0 \\
\end{align}\]
We divide both sides by 5 to have;
\[\Rightarrow 3{{n}^{2}}+5n-1300=0\]
We see that the above equation is quadratic equation in $ n $ which we solve by splitting the middle term $ 5n $ as;
\[\begin{align}
  & \Rightarrow 3{{n}^{2}}-60n+65n-1300=0 \\
 & \Rightarrow 3n\left( n-20 \right)+65\left( n-20 \right)=0 \\
 & \Rightarrow \left( n-20 \right)\left( 3n+65 \right)=0 \\
 & \Rightarrow n=20\text{ or }n=\dfrac{-65}{3} \\
\end{align}\]
Since the number of months cannot be negative we reject the solution $ n=\dfrac{-65}{3} $ and accept the solution as $ n=20 $ . So the man will take 20 months to clear the loan. Hence the correct choice is C.\[\]

Note:
 We note that arithmetic progression; abbreviate d as AP is a type of sequence where the difference between any two consecutive numbers is constant. We can also find the amount he paid in the last month using formula for $ {{n}^{\text{th}}} $ of an AP as $ {{t}_{n}}=a+\left( n-1 \right)d $ .We also note that the question is silent about interest and hence we should not use simple interest formula. We can split the middle term in the quadratic equation from prime factorization of $ 3\times 1300 $ . We can also use quadratic formula but we need to find the square root there,