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Home » A man arranges to 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 instalment.
A man arranges to 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 instalment.
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.7 | Maths | RD Sharma
A man arranges to 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 instalment.

Solution:

As per the question:

There are 40 annual instalments that form an arithmetic series.

Let ‘a‘ be the first instalment

S_{40}=3600, n=40

Using the formula,

\begin{array}{l} S_{n}=n / 2[2 a+(n-1) d] \\ 3600=40 / 2[2 a+(40-1) d] \\ 3600=20[2 a+39 d] \\ 3600 / 20=2 a+39 d \end{array}
2 a+39 d-180=0 \ldots \text { (i) }

Given that,

The sum of first 30 terms is paid and one third of the debt is unpaid.
Therefore, paid amount =2 / 3 \times 3600= ₹ 2400

\mathrm{S}_{\mathrm{n}}=2400, \mathrm{n}=30
Using the formula,
\begin{array}{l} S_{n}=n / 2[2 a+(n-1) d] \\ 2400=30 / 2[2 a+(30-1) d] \\ 2400=15[2 a+29 d] \\ 2400 / 15=2 a+29 d \\ 2 a+29 d-160=0 \ldots \text { (ii) } \end{array}

Let’s now solve eq.(i) and eq.(ii) by substitution method, we obtain
\begin{array}{l} 2 a+39 d=180 \\ 2 a=180-39 d \ldots \text { (iii) } \end{array}

On substituting the value of 2 \mathrm{a} in eq.(ii) we get,
\begin{array}{l} 2 a+29 d-160=0 \\ 180-39 d+29 d-160=0 \\ 20-10 d=0 \\ 10 d=20 \\ d=20 / 10 \\ =2 \end{array}

Substituting the value of d in eq.(iii)
\begin{array}{l} 2 a=180-39 d \\ 2 a=180-39(2) \\ 2 a=180-78 \\ 2 a=102 \\ a=102 / 2 \\ =51 \end{array}

As a result, value of first installment ‘a’ is ₹ 51

i

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