Question

Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month, starting with the first instalment of ₹ 1,000 and he increases the instalment by ₹ 100 every month.

Based on the information given above, answer the following questions:

(i) Find the amount paid by him in the 30^th instalment.   [1]

(ii) If the total number of instalments is 40, what is the amount paid in the last instalment?   [1]

(iii) (a) What amount does he still have to pay after the 30^th instalment?   [2]
OR

(iii) (b) Find the ratio of the tenth instalment to the last instalment.   [2]

Answer 1

Total loan amount = ₹ 1,18,000

First instalment a = 1000

Common difference d = 100

(i) Amount paid by him in 30^th instalment:

a_n = a + (n – 1)d

a₃₀ = 1000 + (30 – 1)100

= 1000 + 29(100)

= 1000 + 2900

a₃₀ = 3900

So, the 30^th instalment is ₹ 3900.

(ii) Amount paid in last instalment:

a_n = a + (n – 1)d

a₄₀ = 1000 + (40 – 1)100

= 1000 + 39(100)

= 1000 + 3900

a₄₀ = 4900

So, the 40^th  (last) instalment is ₹ 4900.

(iii) (a) Amount he still have to pay after 30^th instalment:

First, find the total amount paid in 30 instalment.

`S_n = n/2 [2a + (n - 1)d]`

`S_30 = 30/2 [2(1000) + (30 - 1)100]`

= 15 [2000 + 29(100)]

= 15 [2000 + 2900]

= 15 [4900]

S₃₀ = 73,500

Total amount of loan = 1,18,000

∴ Amount still to pay after 30^th instalment

= Total loan amount – Amount paid till 30 instalment.

= 1,18,000 – 73,500

= 44,500

So, he still has to pay ₹ 44,500 after 30^th instalment.

OR

(iii) (b) Ratio of tenth and last instalment

a₁₀ = 1000 + (10 – 1)d

= 1000 + 9(100)

= 1000 + 900

= a₁₀ = 1900

a10 = 1000 + (40 – 1)100

= 1000 + 39(100)

= 1000 + 3900

= a₄₀ = 4900

∴ `a_10/a_40 = 1900/4900`

= `19/49`