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# Question 65. A man arranges to pay off his debt of Rs 3600 by 40 annual installments 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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This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6
In this question we have been given that a man arranges to pay off his debt of Rs 3600 by 40 annual installments which form an arithmetic series.

When 30 of the installments are paid, he dies leaving one-third of the debt unpaid.

Now we have to find the value of the first installment by using arithmetic progressions properties

CBSE DHANPAT RAI PUBLICATIONS
Understanding CBSE Mathematics
Class :- 10th
Question no 65

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1. Given that the installments are forming an arithmetic series.

Let the first term of the A.P. be a and common difference be d.

Now, Amount of 40 installments = Rs 3600

=> S40 = 3600

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

=> 40[2a + (40 – 1)d] / 2 = 3600

=> 2a + 39d = 180 . . . . (1)

Amount of 30 installments = Rs 3600 – Rs [3600 / 3] = 3600 – 1200 = 2400

=> S30 = 2400

=> 30[2a + (30 – 1)d] / 2 = 2400

=> 2a + 29d = 160 . . . . (2)

On subtracting eq(2) from (1), we get,

=> (2a + 39d) – (2a + 29d) = 180 – 160

=> 10d = 20

=> d = 2

On putting d = 2 in (1), we get,

=> 2a + 39(2) = 180

=> 2a = 180–78

=> 2a = 102

=> a = 51

Hence, value of first installment is 51.

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