Question

A gentleman buys Bank’s cash certificates every year exceeding last year’s purchase by Rs.300. After 20 years, he finds that the total value of the certificate purchased by him is Rs.83,000. Find the value of the certificates purchased by him in the 13th year
A. Rs.4900
B. Rs.6900
C. Rs.1300
D. None of these

Answer 1

Hint: To solve this problem, we need to study and understand the given question keenly. We can solve the given question by using Arithmetic progression. In the given question, we have the difference (d), and the total value of the certificates after 20 years i.e.,\[{{S}_{n}}=83000\]. By using the formula for sum i.e., \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\], from this, we can find that the first year value, then we can simultaneously find the value of the certificates purchased by him in the 13th year by using the formula \[{{a}_{n}}=a+\left( n-1 \right)d\].

Complete step-by-step solution:
Let us solve the given question,
Let the value of the Bank’s cash certificates purchased in the first year be Rs. a.
The difference between the value of the certificates is Rs. 300
So, we can say that common difference, d=300
The total value of the certificate purchased by him =\[{{S}_{n}}=83000\]
Since, it follows Arithmetic progression the total value of the certificate is given by
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Here total number of years =a = 20 years
\[\Rightarrow \dfrac{20}{2}\left[ 2a+19\left( 300 \right) \right]=8300\]
By simplifying
\[\Rightarrow 2a+5700=83000\]
\[\begin{align}
  & \Rightarrow 2a=8300-5700 \\
 & \Rightarrow 2a=2600 \\
 & \Rightarrow a=\frac{2600}{2} \\
 & \therefore a=1300. \\
\end{align}\]
Therefore, the value of the Bank’s cash certificates purchased in the first year will be Rs. 1300.
The value of the certificates purchased by him in nth year= \[a+\left( n-1 \right)d\]
The value of the certificates purchased by him in the 13th year=$1300+(13-1)300= Rs. 4900.$
The correct answer is option(A).

Note: The common mistake done by the students in arithmetic progression is taking the incorrect formula and taking the incorrect values for last and first term. Students get confused about finding the nth term and n. Understanding the problem is also the main criteria.