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
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.