A man arranges to pay off a debt of pounds $3600$ by $40$ annual instalments which form an arithmetic series. When $30$ of the instalments are paid he dies leaving a third of the debt unpaid : find the value of the first instalment.

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Solution

Sum of $40$ installments ${= S}_{40} = 3600$

At the end of $30$ installments debt left $= \frac{1}{3} \times 3600 = 1200$

So the amount paid $= 3600 - 1200 = 2400$

$\Rightarrow S_{30} = 2400$

$S_{n} = \frac{n}{2} {2 a + (n - 1) d} \Rightarrow S_{40} = \frac{40}{2} {2 a + (40 - 1) d} \Rightarrow 3600 = 20 (2 a + 39 d) \Rightarrow 2 a + 39 d = 180....... (i) S_{30} = \frac{30}{2} {2 a + (30 - 1) d} 2400 = 15 (2 a + 29 d) \Rightarrow 2 a + 29 d = 160...... (i i)$
Solving $(i)$ and $(i i)$

$\Rightarrow a = 51, d = 2$
So the first installment is $51$ pounds

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