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.
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
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