Question

Ramesh wants to buy a cell phone. He can buy it by paying $ Rs.15000 $ cash or by making 12 monthly instalments as $ Rs.1800 $ in the $ 1st $ month, $ Rs.1750 $ in $ 2nd $ month, $ Rs.1700 $ in $ 3rd $ month and so on. If he pays the money in instalments, find:
(i.) total amount paid in 12 instalments
(ii.) how much extra he has to pay over and above the cost price

Answer 1

Hint: To solve this question, first we should understand that the question belongs to Arithmetic Progression. We will solve this question with the help of A.P’s formulas of sum of $ n $ terms.

Complete step by step solution:
As we can see that the pattern of question is related with the arithmetic progression.
(i.) Given instalments as per proceeding months are $ Rs.1800,Rs.1750,Rs.1700,..... $
And, we have to find the total amount paid in 12 instalments that is the sum of the amount paid till 12 months.
So, Sum of the 12 instalment as terms:
 $ \therefore {S_n} = \dfrac{n}{2}[a + {a_n}] $
here, $ n $ is the number of total terms which the instalments months;
 $ a $ is the first term which is the first month instalment and
 $ {a_n} $ is the $ nth $ term of the given sequence.
Now, the $ nth $ term of the given sequence is:
 $ {a_n} = a + (n - 1)d $
 $ \Rightarrow {S_n} = \dfrac{n}{2}[a + a + (n - 1)d] $
 $ \Rightarrow {S_n} = \dfrac{n}{2}[2a + (n - 1)d] $
Now,
Difference between the instalment amount, $ d = \text{second instalment} - \text{first instalment} $
or, $ d = 1750 - 1800 = - 50 $
 $ n = 12\, $ (number of total instalment)
 $ a = 1800 $ (first term of the instalment sequence)
So,
Sum of the 12 instalments as terms:
 $
  \therefore {S_n} = \dfrac{n}{2}[2a + (n - 1)d] \\
   \Rightarrow {S_n} = \dfrac{{12}}{2}[2.1800 + (12 - 1)( - 50)] \\
   \Rightarrow {S_n} = 6[3600 + 11( - 50)] \\
   \Rightarrow {S_n} = 6[3600 - 550] \\
   \Rightarrow {S_n} = 6[3050] \\
   \Rightarrow {S_n} = 18300 \;
  $
Hence, the total amount paid in 12 instalments is $ Rs.18300 $ .
(ii.) And, he has to pay extra over and above the cost price is $ Rs.18300 - Rs.15000 = Rs.3300 $
So, the correct answer is “ $ Rs.3300 $ ”.

Note: If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family.