Q) Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs. 1,18,000 by paying every month starting with the first installment of Rs. 1000. If he increases the instalment by Rs. 100 every month , answer the following:
1. The amount paid by him in 30th installment is
a) 3900 b) 3500 c) 3700 d) 3600
2. The amount paid by him in the 30 installments is
a) 37000 b) 73500 c) 75300 d) 75000
3. What amount does he still have to pay offer 30th installment?
a) 45500 b) 49000 c) 44500 d) 54000
4. If total installments are 40 then amount paid in the last installment?
a) 4900 b) 3900 c) 5900 d) 9400
5. The ratio of the 1st installment to the last installment is
a) 1:49 b) 10:49 c) 10:39 d) 39:10
Ans:
1. Starting with 1,000, EMI will be 100 more in next month
∴ AP for this car loan = 1000 + 1100 + 1200 + 1300 + ….
∴ first term a = 1000 and common difference, d = 100
Thus, value of the nth term is, Tn = a + (n – 1) d
∴ T30 = 1000 + (30 – 1) 100
∴ T30 = 1000 + 29 x 100 = 3,900
Therefore, option (a) is correct.
2. The amount paid in 30 installments will be given by the sum of 30 installment values.
Since the sum of n terms of an AP is given by: 
[2 a + (n – 1) d]
Here, we have a = 1000, n = 30 and d = 100
∴ Sum of 30 terms = 
[2 x 1000 + (30 – 1) 100]
= 15 (2000 + 2900) = 15 x 4900 = 73,500
Therefore, option (b) is correct.
3. Total loan amount = 1,18,000
Loan paid till 30th installment = 39,500
Balance loan after 30th installment = 118,000 – 73,500 = 44,500
Therefore, option (c) is correct.
4. Amount paid in 40th installment will be given by value of 40th term.
Let’s put the values of a = 1000, d = 100 and n = 40 in Tn = a + (n – 1) d
∴ T40 = 1000 + (40 – 1) 100
∴ T40 = 1000 + 39 x 100 = 4,900
Therefore, option (a) is correct.
5. 1st installment value = Rs. 1000
40th or last installment value = Rs. 4,900
∴ Ratio of 1st installment to last installment = 
Therefore, option (b) is correct.
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