Question

Ravindra took a loan of ₹3,45,000 from a bank to buy a car and decided to pay back by ₹2,000 at the end of the first month and then increased the instalment amount by ₹200 each month.

Based on the given information, answer the following questions.

1. Find the amount paid by him in the 10th instalment.
2. Find the total amount paid by him in the first 10 instalments.
3. In how many instalments would he clear his total loan?
4. What amount will he be able to clear in his first 45 instalments?

Answer 1

The monthly instalments form an Arithmetic Progression (AP) with:

First instalment, a = ₹2000

Common difference, d = ₹200

Total loan amount, S_n = ₹345000

i. Amount paid in the 10th instalment:

Formula: \[T_n = a + (n - 1)d\]

Calculation: \[T_{10} = 2000 + (10 - 1) \times 200\]

\[T_{10} = 2000 + 9 \times 200\]

\[= 2000 + 1800\]

\[= ₹3800\]

He paid ₹3,800 in the 10th instalment.

ii. Total amount paid in the first 10 instalments:

Formula: \[S_n = \frac{n}{2}[2a + (n - 1)d]\]

Calculation: \[S_{10} = \frac{10}{2}[2(2000) + (10 - 1)200]\]

\[S_{10} = 5 \times [4000 + 1800]\]

\[= 5 \times 5800\]

\[= ₹29000\]

Total amount paid in first 10 instalments is ₹29,000.

iii. Number of instalments to clear the total loan:

Given: \[S_n = 345000\] 

\[\frac{n}{2}[2(2000) + (n - 1)200] = 345000\] 

\[\frac{n}{2}[4000 + 200n - 200] = 345000\] 

\[\frac{n}{2}[200n + 3800] = 345000\] 

\[n[100n + 1900] = 345000\] 

\[100n^2 + 1900n - 345000 = 0\] 

Dividing by 100: \[n^2 + 19n - 3450 = 0\] 

Factoring the quadratic equation: \[n^2 + 69n - 50n - 3450 = 0\] 

\[n(n + 69) - 50(n + 69) = 0\] 

\[(n + 69)(n - 50) = 0\] 

\[n = -69 \quad \text{or} \quad n = 50\] 

Since the number of instalments must be positive, n = 50.

He will clear his total loan in 50 instalments.

iv. Amount cleared in his first 45 instalments:

Calculation for n = 45: \[S_{45} = \frac{45}{2}[2(2000) + (45 - 1)200]\]

\[S_{45} = \frac{45}{2}[4000 + 44 \times 200]\] 

\[S_{45} = \frac{45}{2}[4000 + 8800] = \frac{45}{2} \times 12800\] 

\[S_{45} = 45 \times 6400 = ₹288000\]

He will be able to clear ₹2,88,000 in 45 instalments.