Question

A gentleman buys every year Bank’s cash certificates of values exceeding last year’s purchase by Rs.300 .After 20 years, he finds that the total value of the certificates purchased by him is Rs.83000 . Find the value of certificates purchased by him in the 13th year.
A)Rs.4900 B) Rs.6900 C) Rs.3900 D)None of these

Answer 1

Hint-The difference of the value of the certificates every year is 300 which means it follows Arithmetic Progression. Find the value of certificates purchased in 1st year using formula Sn$= \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ .Then find the value of certificates in 13th year by using nth term formula.

Complete step-by-step answer:
Given, The difference between the values of the certificates is Rs.300 And after 20 years, the total value of certificates is Rs.83,000.Let, the value of certificates purchased by the gentleman in 1st year be ‘a’. Given d=300 and sum after 20 years=83000 .Since it follows A.P. then using formula, Sn$= \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ $\Rightarrow 8300 = \dfrac{{20}}{2}\left[ {2a + \left( {20 - 1} \right)300} \right] = 10\left[ {2a + 19\left( {300} \right)} \right]$
$8300 = 2a + 5700 \Rightarrow 2a = 8300 - 5700 = 2600 \\\ a = \dfrac{{2600}}{2} = 1300 \\\$
Then the value of certificates purchased by him in nth year=$a + \left( {n - 1} \right)d$
Here, a=1300 and n=13 and d=300
Thus, the value of his certificates in 13th year =$1300 + \left( {13 - 1} \right)300 = 1300 + 3600 = 4900$
Hence the correct answer is A

Note: To solve this type of Question, first we have to see that the difference between two consecutive terms is always the same only then will the formulas of A.P. can be applied.