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Find the value of the first installment. A man arranges to pay off a debt of Rs. 3600 in 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid.

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Rs. 49

Rs. 51

Rs. 53

Rs. 55

answer is B.

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Detailed Solution

Given, the debt of

Rs . 3600

is to be paid in

40

annual installments.
The sum of 40 years is termed as,

S_{40} = 3600 . . . . . . . . (i)

After

30

installments one-third of the debt is unpaid,

\Rightarrow S_{30} = 3600 - \frac{1}{3} of 3600

\Rightarrow S_{30} = 3600 - \frac{1}{3} \times 3600

\Rightarrow S_{30} = 3600 - 1200

{\Rightarrow S}_{30} = 2400 . . . . . . . . (ii)

The sum of terms be

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

Substitute the values,

\Rightarrow S_{40} = \frac{40}{2} [2 a + 39 d]

\Rightarrow \frac{40}{2} [2 a + 39 d] = 3600

\Rightarrow 2 a + 39 d = 180 . . . . . (iii)

From the 2nd equation,

S_{30} = 2400

\Rightarrow \frac{30}{2} [2 a + 29 d] = 2400

\Rightarrow 2 a + 29 d = 160 . . . . . . (iv)

Solve equation iii and iv,

\Rightarrow (2 a + 39 d) - (2 a + 29 d) = 180 - 160

\Rightarrow 10 d = 20

\Rightarrow d = 2

Substitute the value in equation iii to get the first term value,

\Rightarrow 2 a = 180 - 39 d

\Rightarrow 2 a = 180 - (39 \times 2)

\Rightarrow 2 a = 180 - 78

\Rightarrow 2 a = 102

\Rightarrow a = \frac{102}{2}

\Rightarrow a = 51

Hence, the value of the first installment is

Rs . 51

. Therefore, option 2 is correct.

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