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A man arranges to pay off a debt of Rs. $3600$ in $40$ annual instalments which form an arithmetic series. When $30$ of the instalments are paid he dies leaving one-third of the debt unpaid. Find the value of the first instalment.
Given: A man arranges to pay off a debt of Rs. $3600$ in $40$ annual instalments which form an arithmetic series. When $30$ of the instalments are paid he dies leaving one-third of the debt unpaid.
To do: To find the value of the first instalment.
Solution:
According to the question,
The total amount of debit to be paid in $40$ installment $= Rs.\ 3600$
After $30$ installment one-third of his debit is left unpaid. This means that he paid two third of the payment.
So, the amount he paid in 30 installments $S_{30}=( 3600)-\frac{1}{3}( 3600)$
$\Rightarrow S_{30}= 3600- 1200$
$\Rightarrow S_{30}= 2400$
Let us take the first installment as $a$ and common difference as $d$.
So, using the formula for the sum of $n$ terms of an A.P.
$Sn=\frac{n}{2}[2a+( n-1)d]$
Let us find $a$ and $d$, for $30$ installments.
$S_{30}=\frac{30}{2}[2a+(30-1)d]$
$2400=15[2a+( 29)d]$
$\frac{2400}{15}=2a+29d$
$160=2a+29d$
$a=\frac{160-29d}{2} .....( 1)$
Similarly, we find $a$ and $d$ for $40$ installment.
$S_{40}=\frac{40}{2}[2a+( 40-1)d]$
$\Rightarrow 3600=20( 2a+( 39)d)$
$\Rightarrow \frac{3600}{20}=2a+39d$
$a=\frac{180-39d}{2}\ \ ........( 2)$
subtracting $( 1)$ from $( 2)$, we get
$a-a=\frac{( 180-39d)}{2}-\frac{( 160-29d)}{2}$
$\Rightarrow \frac{180-39d-160+29d}{2}=0$
$\Rightarrow 20-10d=0$
$\Rightarrow 10d = 20$
$\Rightarrow d=\frac{20}{10}$
$\Rightarrow d = 2$
Subtracting the value of $d$ in $( 1)$, we get,
$a=\frac{160-29(2)}{2}$
$=\frac{160-58}{2}$
$=\frac{102}{2}$
$= Rs\ 51$
Thus, First installment is $Rs.\ 51$.
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