A man arranges too pay off a debt of $₹ 3600$ by $40$ 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.

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Solution

Total amount of debt to be paid in $40$ instalments $= ₹ 3600$
After $30$ instalments, one-third of his debt is unpaid which means that two-third of his payment is done.
Therefore, the amount he paid in $30$ instalments $= ₹ (\frac{2}{3} \times 3600)$
$= ₹ (2 \times 1200) = ₹ 2400$
Let $a$ be the first instalment and $d$ be the common difference between them.
Step $1$ : Find a using S_nfor $30$ instalments.
We know from the formula of AP that,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Substituting the values of $a$ and $d$ ,
$S_{30} = \frac{30}{2} [2 a + (30 - 1) d] \Rightarrow 2400 = 15 [2 a + 29 d] \Rightarrow [2 a + 29 d] = \frac{2400}{15} \Rightarrow [2 a + 29 d] = 160 \Rightarrow 2 a = 160 - 29 d \Rightarrow a = \frac{160 - 29 d}{2} . . . . . . . . . . . . . . . [1]$
Step $2$ : Find a using S_nfor $40$ instalments.
Similarly, $a$ and $d$ are calculated for $40$ instalments
$S_{40} = \frac{40}{2} [2 a + (40 - 1) d] \Rightarrow 3600 = 20 [2 a + 39 d] \Rightarrow [2 a + 39 d] = 180 \Rightarrow 2 a = 180 - 39 d \Rightarrow a = \frac{180 - 39 d}{2} . . . . . . . . . . . . . [2]$
Step 3 : Find a and d using above steps.
Subtracting $e q . [1]$ from $e q . [2]$ , we get
$a - a = (\frac{180 - 39 d}{2}) - (\frac{160 - 29 d}{2}) \Rightarrow 0 = \frac{180 - 39 d - 160 + 29 d}{2} \Rightarrow 0 = \frac{20 - 10 d}{2} \Rightarrow 20 - 10 d = 0 \Rightarrow 10 d = 20 \Rightarrow d = \frac{20}{10} \Rightarrow d = 2$
Substituting the value of $d$ ,
$a = \frac{160 - 29 \times (2)}{2} \Rightarrow a = \frac{160 - 58}{2} \Rightarrow a = \frac{102}{2} \Rightarrow a = 51$
Hence, $₹ 51$ was paid in the first instalment.

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When $30$ of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.