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 of1,000 and he increases the instalment by 100 every month.
Question: 1
Find the amount paid by him in the \( 30^{\text{th}} \) instalment.
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Always identify 'a' and 'd' correctly from the word problem. The "starting amount" is 'a' and the "monthly increase" is 'd'.
Step 1: Understanding the Concept:
The monthly instalments form an Arithmetic Progression (A.P.) because there is a constant increase in the payment every month. Step 2: Key Formula or Approach:
The \( n^{\text{th}} \) term of an A.P. is given by the formula:
\[ a_n = a + (n - 1)d \]
Where:
\( a = \) first term (Rs 1,000)
\( d = \) common difference (Rs 100)
\( n = 30 \) for the \( 30^{\text{th}} \) instalment. Step 3: Detailed Explanation:
Substitute the values into the formula:
\[ a_{30} = 1000 + (30 - 1) \times 100 \]
\[ a_{30} = 1000 + 29 \times 100 \]
\[ a_{30} = 1000 + 2900 \]
\[ a_{30} = 3900 \] Step 4: Final Answer:
The amount paid in the \( 30^{\text{th}} \) instalment is Rs 3,900.
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Question: 2
If the total number of instalments is 40, what is the amount paid in the last instalment ?
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The "last instalment" is simply the term corresponding to the total number of periods given in the problem.
Step 1: Understanding the Concept:
The last instalment refers to the \( 40^{\text{th}} \) term of the A.P. sequence. Step 2: Detailed Explanation:
Here, \( n = 40 \), \( a = 1000 \), and \( d = 100 \).
Using the formula \( a_n = a + (n - 1)d \):
\[ a_{40} = 1000 + (40 - 1) \times 100 \]
\[ a_{40} = 1000 + 39 \times 100 \]
\[ a_{40} = 1000 + 3900 \]
\[ a_{40} = 4900 \] Step 3: Final Answer:
The amount paid in the last (\( 40^{\text{th}} \)) instalment is Rs 4,900.
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Question: 3
What amount does he still have to pay after the \( 30^{\text{th}} \) instalment ?
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When asked for "remaining amount", always calculate the sum paid so far using the \( S_n \) formula, not just the \( n^{\text{th}} \) term.
Step 1: Understanding the Concept:
To find the remaining amount, we first calculate the sum of the first 30 instalments and subtract it from the total loan amount. Step 2: Key Formula or Approach:
Sum of first \( n \) terms of an A.P.:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \] Step 3: Detailed Explanation:
Total loan = Rs 1,18,000.
Sum of first 30 instalments (\( S_{30} \)):
\[ S_{30} = \frac{30}{2} [2(1000) + (30 - 1) \times 100] \]
\[ S_{30} = 15 [2000 + 2900] \]
\[ S_{30} = 15 \times 4900 = 73,500 \]
Amount remaining = Total loan \( - S_{30} \)
\[ \text{Remaining} = 1,18,000 - 73,500 = 44,500 \] Step 4: Final Answer:
The amount left to pay after the \( 30^{\text{th}} \) instalment is Rs 44,500.
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Question: 4
Find the ratio of the tenth instalment to the last instalment.
Show Hint
Always simplify the ratio by canceling out common factors (like zeros) to reach the simplest integer form.
Step 1: Understanding the Concept:
Ratio is the comparison of two quantities by division. We need the values of the \( 10^{\text{th}} \) and \( 40^{\text{th}} \) instalments. Step 2: Detailed Explanation:
Calculate \( 10^{\text{th}} \) instalment (\( a_{10} \)):
\[ a_{10} = 1000 + (10 - 1) \times 100 = 1000 + 900 = 1,900 \]
Calculate last (\( 40^{\text{th}} \)) instalment (\( a_{40} \)):
From part (ii), \( a_{40} = 4,900 \).
Ratio = \( a_{10} : a_{40} \)
\[ \text{Ratio} = \frac{1900}{4900} = \frac{19}{49} \] Step 3: Final Answer:
The ratio is 19 : 49.
