Question

A man arranges to pay off a debt of Rs.3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one – third of the debt unpaid. Find the value of the first installment.

Answer 1

First of all, use the formula of sum of n terms of an A.P, which is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ to find the sum of 40 installments by substituting n = 40 and ${{S}_{n}}=3600$. Then find the sum of the last 10 installments by using ${{S}_{40}}-{{S}_{30}}=1200$. By solving these two equations, get the value of ‘a’ that is the first installment.

Complete step-by-step answer:
We are given that a man arranges to pay a debt of Rs.3600 by paying 40 annual installations which are in arithmetic progression. It is also given that he dies after paying 30 installments leaving one – third of the total debt unpaid. We have to find the value of the first installment.
Let us consider the first installment paid by the man to be Rs. $a$.
Since we know that the installments are in A.P. Therefore, let us assume that next installments are as follows:
$a,a+d,a+2d,a+3d,a+4d......$
where ‘a’ is the first term and ‘d’ is the common difference of A.P.
We know that the general term of A.P is ${{a}_{n}}=a+\left( n-1 \right)d$.
Since we know that there are total of 40 installments, therefore,
We get 40th installment as
${{a}_{40}}=a+\left( 40-1 \right)d$
$=a+39d$
Hence, we get the 1st to 40th installments as follows
$a,a+d,a+2d,a+3d.......a+39d$
We know that the sum of n terms of A.P is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
We are given that the sum of these 40 installments is Rs.3600.
Hence, we get,
${{S}_{40}}=\dfrac{40}{2}\left[ 2a+\left( 40-1 \right)d \right]=3600$
By simplifying the above equation, we get,
$20\left( 2a+39d \right)=3600$
By dividing 20 on both sides, we get,
$2a+39d=180.....\left( i \right)$
Now, we are given that he died after paying the first 30 installments leaving one-third of the debt unpaid. That means the sum of the last 10 installments is equal to one-third of debt. We can write it as
Sum of last 10 installments $=\dfrac{1}{3}[\text{Total Debt}]$
Since we know that the total debt is Rs.3600. By substituting its value in the above equation, we get,
Sum of last 10 installments $=\dfrac{3600}{3}=1200.....\left( ii \right)$
Now, the sum of last 10 installments = (Sum of total 40 installments) – (Sum of first 30 installments)
$={{S}_{40}}-{{S}_{30}}$
We know that the sum of the total of 40 installments $={{S}_{40}}=Rs.3600$
And the sum of the first 30 installments $={{S}_{30}}=\dfrac{30}{2}\left( 2a+\left( 30-1 \right)d \right)$
We get, ${{S}_{30}}=15\left( 2a+29d \right)$
By substituting the value of ${{S}_{40}}$ and ${{S}_{30}}$, we get,
Sum of last 10 installments $=3600-15\left( 2a+29d \right)$
By substituting the value of the sum of the last 10 installments in equation (ii), we get
$3600-15\left( 2a+29d \right)=1200$
Or, $15\left( 2a+29d \right)=3600-1200=2400$
By dividing 15 on both sides, we get,
$\left( 2a+29d \right)=160....\left( iii \right)$
By substituting equation (iii) from (ii), we get,
$\left( 2a+39d \right)-\left( 2a+29d \right)=180-160$
We get, $\left( 39d-29d \right)=20$
Or, $10d=20$
By dividing 10 on both sides, we get,
$d=2$
By substituting the value of d = 2 in equation (i), we get,
$2a+39\times 2=180$
Or, $2a+78=180$
$\Rightarrow 2a=180-78$
$\Rightarrow 2a=102$
By dividing 2 on both sides, we get,
$a=51$
As we have assumed that the first installment is Rs a, therefore we get the value of the first installment as Rs. 51.

Note: Some students make this mistake of finding the sum of the last 10 installments by using the formula of ${{S}_{10}}$ that is $\dfrac{10}{2}\left( 2a+\left( 10-1 \right)d \right)$ but this is wrong. They must note that ${{S}_{10}}$ is the formula to find the sum of the first 10 installments and sum of the last 10 installments i.e. $\left( {{S}_{40}}-{{S}_{30}} \right)$. In general, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ is the formula of sum of n terms of A.P starting from first term that is ‘a’ and this must be taken care of.