Q: 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 30th instalment.
(ii)  If the total number of instalments is 40, what is the amount paid in the last instalment ?
(iii) What amount does he still have to pay after the 30th instalment?
(iv) Find the ratio of the tenth instalment to the last instalment.

[Q 38 – 30/2/2 – CBSE 2026 Question Paper]

Ans:

We are given the following values:

Total loan = ₹ 1,18,000

First instalment = ₹1,000

Each instalment increases by ₹100 every month

Therefore, this forms an Arithmetic Progression (AP):

Here, first term, a = 1000

and Common difference, d = 100

(i) Amount paid in the 30th installment: 

We can get this amount by finding the value of 30th term.

Since the value of nth term is given by, T_n = a + (n – 1) × d

∴ T₃₀ = 1000 + (30 – 1) × 100

= 1000 + 2900 = ₹ 3,900

Therefore, amount paid in the 30th instalment is ₹ 3,900.

(ii) Amount paid in the last instalment:

Since 40th instalment is the last instalment, we can get this amount by finding the value of 40th term.

Again by nth term formula, T_n = a + (n – 1) × d

∴ T₄₀ = 1000 + (40 – 1) × 100

= 1000 + 3900 = ₹ 4,900

Therefore, amount paid in the last instalment is ₹ 4,900.

(iii) Balance amount to be paid after 30 instalments:

Here, we can see that amount is being paid right from 1st to 30th instalments.

So if we find out total amount paid in these first 30 instalments, we can find out the amount yet to be paid.

Step 1: Let’s calculate the total amount paid in 30 instalments:

Since sum of first n terms of an AP is given by: S_n = [2 a + (n – 1) d]

∴ Sum of first 30 terms, S₃₀= [2 (1000) + (30 – 1) (100)]

= 15 x (2000 + 2900) = 15 × 4900

= ₹ 73,500

Step 2: It is given that total loan amount to be paid is ₹ 1,18,000

Balance amount = 1,18,000 – 73,500

= 44,500

Therefore, after paying 30 instalments, ₹ 44,500 are yet to be paid.

(iv) Ratio of 10^th instalment to last instalment:

To calculate this ratio, first we need to have value of 10^th and last instalments.

Step 1: First we calculate value of 10^th instalment by n^th term formula:

∵ T_n = a + (n – 1) × d

∴ T₁₀ = 1000 + (10 – 1) × 100

= 1000 + 900 = 1,900

∴  T₁₀ = ₹ 1,900

Step 2: Next, we need to find value of last instalment

and for this we need to find which term is last instalment

Since, it is given that total loan amount is ₹ 1,18,000

Let’s consider n installments are paid

∵ S_n = [2 a + (n – 1) d]

∴ 118000 =  [2 (1000) + (n – 1) (100)]

∴ 1180  = [2 (10) + (n – 1)]

∴ 1180 x 2 = n (20 + n – 1)

∴ 2360 = n (19 + n)

∴ n ² +19 n – 2360 = 0

∴ n ² + 59 n – 40 n + 2360 = 0   (by mid term splitting)

∴ n ( n + 59) – 40 (n – 59) = 0

∴ (n + 59) (n – 40) =  0

∴ n = – 59 and n = 40

Since value of n can not be negative, hence we reject n = – 59

and accept n = 40

40^th instalment is the last instalment

Step 3: Now, we know that 40^th instalment is the last instalment

and its value already calculated in part (ii),

∴ T₄₀ = ₹ 4,900

Step 4: ∴ Ratio =

∴ T₁₀ : T₄₀ = 19 : 49

Therefore, the ratio of 10^th and the last (40^th) instalment is 19:49.

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