Reshma wanted to save at least ₹ 6,500 for sending her daughter to school next year (after 12 months). She saved ₹ 450 in the first month and raised her savings by ₹ 20 every next month. How much will she be able to save in next 12 months. ? Will she be able to send her daughter to the school next year ?
What value is reflected in this question ?

Answer

To solve the problem step by step, we will calculate how much Reshma can save over the next 12 months and determine if she can send her daughter to school.

### Step 1: Identify the savings pattern
Reshma saves ₹450 in the first month and increases her savings by ₹20 every subsequent month.

### Step 2: Calculate the savings for each month
- Month 1: ₹450
- Month 2: ₹450 + ₹20 = ₹470
- Month 3: ₹470 + ₹20 = ₹490
- Month 4: ₹490 + ₹20 = ₹510
- Month 5: ₹510 + ₹20 = ₹530
- Month 6: ₹530 + ₹20 = ₹550
- Month 7: ₹550 + ₹20 = ₹570
- Month 8: ₹570 + ₹20 = ₹590
- Month 9: ₹590 + ₹20 = ₹610
- Month 10: ₹610 + ₹20 = ₹630
- Month 11: ₹630 + ₹20 = ₹650
- Month 12: ₹650 + ₹20 = ₹670

### Step 3: Sum the savings over 12 months
The total savings can be calculated using the formula for the sum of an arithmetic progression (AP):
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Where:
- \( n \) = number of terms (12 months)
- \( a \) = first term (₹450)
- \( d \) = common difference (₹20)

Substituting the values:
\[ S_{12} = \frac{12}{2} \times (2 \times 450 + (12 - 1) \times 20) \]
\[ S_{12} = 6 \times (900 + 220) \]
\[ S_{12} = 6 \times 1120 \]
\[ S_{12} = 6720 \]

### Step 4: Compare savings with the target amount
Reshma needs at least ₹6,500 to send her daughter to school. Since she will save ₹6,720 in total, she will be able to send her daughter to school.

### Conclusion
Yes, Reshma will be able to send her daughter to school next year as she will save ₹6,720, which is more than the required ₹6,500.

### Value Reflected
The value reflected in this question is the importance of education for girls and the commitment of parents to save for their children's education.

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