Question

A man purchased LCD TV for Rs. 32,500. He paid Rs. 200 initially and increased the payments by Rs. 150 every month. How many months did he take to complete payment?

Answer 1

Hint: It is given that the price at which the man purchased LCD TV is equal to Rs. 32,500. The man paid Rs. 200 initially. After paying Rs. 200 initially, the man increases the payments every month by Rs. 150. We can observe that the payment for every next month is increasing in arithmetic progression. The first term of the AP is Rs. 200 whereas the common difference is Rs. 150. Let us assume that after n months the man completes the payment. Now, calculate the total amount of money that the man paid in n months using the formula, \[\dfrac{n}{2}\left[ 2First \space term+\left( n-1 \right)common \space difference \right]\] . The total amount of money that the man paid in n months is exactly equal to the price at which the man purchased LCD TV. Now, solve it further and get the value of n.

Complete step-by-step answer:
According to the question, it is given that a man purchased an LCD TV for Rs. 32,500. He paid Rs. 200 initially and increased the payments by Rs. 150 every month.
The price at which the man purchased LCD TV = Rs. 32,500 …………………………..(1)
The first payment that the man pays = Rs. 200 ……………………………..(2)
It is given that every subsequent payment is Rs. 150 more than the previous payment. In the next month, the man pays Rs. 150 more than that of Rs. 200, \[Rs.150+Rs.200=Rs.350\] . Similarly, In the next to next month, the man pays Rs. 150 more than that of Rs. 350, \[Rs.350+Rs.150=Rs.500\] . We can observe that the payment follows the arithmetic progression where the first term is Rs. 200 and the common difference is Rs. 150.
The first term = Rs. 200 ……………………………..(3)
The common difference = Rs. 150 …………………………………(4)
Let us assume that after n months the man completes the payment.
We know the formula for the summation of n terms in an A.P, \[\dfrac{n}{2}\left[ 2First \space term+\left( n-1 \right)common \space difference \right]\] ………………………….(5)
The total amount of money that he paid in n months = \[\dfrac{n}{2}\left[ 2First \space term+\left( n-1 \right)common \space difference \right]\] …………………………………….(6)
Now, from equation (3), equation (4), and equation (6), we get
The total amount of money that he paid in n months =\[\dfrac{n}{2}\left[ 2\times 200+\left( n-1 \right)150 \right]\] = \[\dfrac{n}{2}\left[ 150n+250 \right]\] ……………………………………………..(7)
The total amount of money that the man paid in n months is exactly equal to the price at which the man purchased LCD TV.
Now, on comparing (1) and equation (7), we get
\[\begin{align}
  & \Rightarrow 32500=\dfrac{n}{2}\left[ 150n+250 \right] \\
 & \Rightarrow 32500=\dfrac{50n}{2}\left[ 3n+5 \right] \\
 & \Rightarrow 650\times 2=3{{n}^{2}}+5n \\
 & \Rightarrow 3{{n}^{2}}+5n-1300=0 \\
 & \Rightarrow 3{{n}^{2}}+65n-60n-1300=0 \\
 & \Rightarrow n\left( 3n+65 \right)-20\left( 3n+65 \right)=0 \\
 & \Rightarrow \left( n-20 \right)\left( 3n+65 \right)=0 \\
\end{align}\]
So, \[n=20\] or \[n=\dfrac{-65}{3}\] .
The value of n cannot be negative. So, \[n=\dfrac{-65}{3}\] is not possible.
The value of n must be positive. So, \[n=20\] is possible.
Therefore, after 20 months the man completes the payment of the LCD TV.



Note:In this question, one might think that after paying Rs. 200 initially, the man pays Rs. 150 every month. This is wrong because after paying Rs. 200 initially, the man increases the payments every month by Rs. 150. It means that in the next month the man pays Rs. 350 and in the next to next month the man pays Rs. 500. Now, we can see the pattern of the payments that the man pays. The amount of payment is increasing in the Arithmetic progression form.