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प्रश्न
| 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.
- Find the amount paid by him in the 10th instalment.
- Find the total amount paid by him in the first 10 instalments.
- In how many instalments would he clear his total loan?
- What amount will he be able to clear in his first 45 instalments?
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उत्तर
The monthly instalments form an Arithmetic Progression (AP) with:
First instalment, a = ₹2000
Common difference, d = ₹200
Total loan amount, Sn = ₹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.
