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Arithmetic Progressions
EXERCISE 5.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Ramkali saved ` 5 in the first week of a year and then increased her weekly savings by ` 1.75. If in the nth week, her weekly savings become ` 20.75, find n.

Step-by-Step Solution

Key Concept: Use the nth term formula of an arithmetic progression: \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
1. Identify the given data:
- First week saving (first term) \(a = \text{` }5\).
- Weekly increase (common difference) \(d = \text{` }1.75\).
- Saving in the \(n\)th week (nth term) \(a_n = \text{` }20.75\).

2. Write the nth term formula for an AP:
$$a_n = a + (n-1)d$$

3. Substitute the known values:
$$20.75 = 5 + (n-1)\times 1.75$$

4. Isolate \((n-1)\):
$$20.75 - 5 = (n-1)\times 1.75$$
$$15.75 = (n-1)\times 1.75$$

5. Divide both sides by \(1.75\):
$$n-1 = \frac{15.75}{1.75}$$
Since \(1.75 \times 9 = 15.75\), we get
$$n-1 = 9$$

6. Solve for \(n\):
$$n = 9 + 1 = 10$$

Therefore, Ramkali reaches a weekly saving of `20.75 in the 10th week.

Correct Answer: 10
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