Asif purchased a car for 360000. The Amount is payable in forty monthly installments which are in arithmetic progression. After paying thirsty installments one third of the amount would remain unpaid. The amount which Asif would be required to pay as the 35th installment is
Let the first installment be a and the common difference between the installments be d.
The arithmetic progression for the installments can be represented as: a, a+d, a+2d, a+3d, ..., a+39d.
The total cost of the car is given as 360000.
The sum of n terms of an arithmetic progression is given by the formula: Sn = (n/2)(2a + (n-1)d).
Since there are 40 installments, we can represent this as: Sn = (40/2)(2a + (40-1)d).
After paying thirty installments, one third of the amount would remain unpaid. This means that the remaining amount to be paid after thirty installments is (2/3)(360000) = 240000.
So the sum of the first thirty installments is: S30 = (30/2)(2a + (30-1)d) = 240000.
We can simplify this to: 15(2a + 29d) = 240000.
Simplifying further: 2a + 29d = 16000.
We can use this equation to solve for a in terms of d: a = (16000 - 29d)/2.
Now we need to find the value of the 35th installment, which can be represented as: a + 34d.
Substituting the value of a we found earlier: (16000 - 29d)/2 + 34d.
Multiplying through by 2 to remove the fraction: 16000 - 29d + 68d.
Combine like terms: 39d.
Therefore, the value of the 35th installment is 39d.
To find the amount Asif would be required to pay as the 35th installment, we need to find out the common difference and the first term of the arithmetic progression.
Let's call the first term of the arithmetic progression 'a' and the common difference 'd'.
The formula to find the nth term of an arithmetic progression is:
nth term = a + (n - 1)d
We know that after paying thirty installments, one third of the amount is still unpaid. So, the remaining amount can be calculated as:
Remaining amount = 360000 / 3 = 120000
Since there are forty monthly installments, the 40th installment would be the last installment. Therefore, the remaining amount after thirty installments should be equal to the sum of the remaining ten installments, excluding the 40th installment.
Using the formula for the sum of an arithmetic progression, we can find the sum of ten terms:
Sum = (n/2)(2a + (n-1)d)
where n = 10
Substituting the values into the formula:
120000 = (10/2)(2a + (10-1)d)
24000 = 10(2a + 9d)
24000 = 20a + 90d
Similarly, using the formula for the sum of thirty terms, we can find the sum of thirty installments:
Sum = (n/2)(2a + (n-1)d)
where n = 30
Substituting the values into the formula:
360000 - 120000 = (30/2)(2a + (30-1)d)
240000 = 15(2a + 29d)
16000 = 2a + 29d (equation 1)
Now, we have two equations:
24000 = 20a + 90d (equation 2)
16000 = 2a + 29d (equation 1)
We can solve these two equations to find the values of 'a' and 'd'.
Multiplying equation 1 by 10, we get:
160000 = 20a + 290d (equation 1 multiplied by 10)
240000 = 20a + 90d (equation 2)
Subtracting equation 2 from the equation 1 multiplied by 10, we get:
160000 - 240000 = 290d - 90d
-80000 = 200d
d = -80000/200
d = -400
Substituting the value of 'd' into equation 2, we can find 'a':
24000 = 20a + 90(-400)
24000 = 20a - 36000
20a = 24000 + 36000
20a = 60000
a = 60000/20
a = 3000
So, the first term of the arithmetic progression is 3000 and the common difference is -400.
To find the 35th installment, we now use the formula for the nth term of an arithmetic progression:
35th installment = a + (n - 1)d
35th installment = 3000 + (35 - 1)(-400)
35th installment = 3000 + 34(-400)
35th installment = 3000 - 13600
35th installment = -10600
Therefore, Asif would be required to pay -10600 as the 35th installment.
To find the amount which Asif would be required to pay as the 35th installment, we need to find the common difference and the first term of the arithmetic progression.
Let's assume that the first term of the arithmetic progression is 'a' and the common difference is 'd'.
Given that Asif purchased a car for 360000 and the amount is payable in forty monthly installments, we can write the sum of the arithmetic progression as follows:
S = (n/2) * [2a + (n-1)d]
where S is the total amount to be paid, n is the number of installments, a is the first term, and d is the common difference.
In this case, n = 40, S = 360000, and t = 35.
After paying thirty installments, one-third of the amount would remain unpaid. This means that two-thirds of the total amount has been paid after 30 installments. Therefore, we can write:
(2/3) * S = (30/2) * [2a + (30-1)d]
Now we can solve these two equations simultaneously to find the values of 'a' and 'd'.
Equation 1: (40/2) * [2a + (40-1)d] = 360000
Equation 2: (2/3) * 360000 = (30/2) * [2a + (30-1)d]
Simplifying Equation 2: 240000 = (15/2) * [2a + 29d]
Now we have a system of two equations with two variables (a and d). We can solve these equations to find the values of 'a' and 'd'.
Solving equations simultaneously:
From Equation 2, we can isolate 'a':
240000 = (15/2) * [2a + 29d]
(2/15) * 240000 = 2a + 29d
a = (8/15) * 240000 - (29/15)d
a = 128000 - (29/15)d
Substituting this value of 'a' in Equation 1:
(40/2) * [2(128000 - (29/15)d) + (40-1)d] = 360000
4 * [256000 - (29/15)d + 39d] = 360000
1024000 - 7.6d + 156d = 360000
-7.6d + 156d = 360000 - 1024000
148.4d = -664000
d = -664000/148.4
d = -4477.18 (approx)
Now, substituting the value of 'd' in the value of 'a':
a = 128000 - (29/15)d
a = 128000 - (29/15)(-4477.18)
a = 128000 + 86223.33
a = 214223.33 (approx)
We have found the values of 'a' (first term) and 'd' (common difference). Now, we can find the 35th installment using the formula:
35th installment = a + (35-1)d
35th installment = 214223.33 + (35-1)(-4477.18)
35th installment = 214223.33 + 34(-4477.18)
35th installment = 214223.33 - 152167.12
35th installment = 62056.21
Therefore, the amount which Asif would be required to pay as the 35th installment is approximately 62056.21.