A man arranges to pay off a debt of $R s 3600$ by $40$ annual installment 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.

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Solution

$S_{40} = 3600, S_{30} = 3600 - \frac{1}{3} [3600] = 2400$ ,
because after paying $30$ instalments $\frac{1}{3}$ rd of debt is still left. The given two equations give $2 a + 39 d = 180$ and $2 a + 29 d = 160$ . Solving them, we get $d = 2$ and $a = 51$ , i.e., the value of first instalment is Rs. $51$ .

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Q. A man arranges to pay off a debt of Rs.

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annual installments which form an arithmetic series. When

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of the installments are paid, he dies leaving one third of the debt unpaid, find the value of the first installment.

A man arranges to pay off a debt of Rs. 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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A man arranges to pay off a debt of Rs 3600 by 40 annual installment 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.

S40=3600,S30=3600−13[3600]=2400,because after paying 30 instalments 13rd of debt is still left. The given two equations give 2a+39d=180 and 2a+29d=160. Solving them, we get d=2 and a=51, i.e., the value of first instalment is Rs. 51.

A man arranges to pay off a debt of $R s 3600$ by $40$ annual installment 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.