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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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