To solve this problem, we will focus on understanding the pattern of savings and then calculating the total amount saved over a decade.
1. Identify the pattern of savings:
- In the first month, the person saves Rs. 200.
- In the second month, the saving increases by Rs. 100, making it Rs. 300.
- In the third month, it further increases by Rs. 100, making it Rs. 400.
- This pattern continues, with the monthly saving increasing by Rs. 100 every month.
2. Formulate the sequence:
- The savings form an arithmetic series where the first term [tex]\( a \)[/tex] is Rs. 200 and the common difference [tex]\( d \)[/tex] is Rs. 100.
3. Determine the number of months in a decade:
- A decade is 10 years, and there are 12 months in a year.
- Therefore, the total number of months [tex]\( n \)[/tex] in a decade is [tex]\( 10 \times 12 = 120 \)[/tex] months.
4. Use the formula for the sum of an arithmetic series:
- The formula to find the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series is:
[tex]\[
S_n = \frac{n}{2} \times (2a + (n - 1)d)
\][/tex]
where:
- [tex]\( n \)[/tex] is the number of terms (120 months),
- [tex]\( a \)[/tex] is the first term (Rs. 200),
- [tex]\( d \)[/tex] is the common difference (Rs. 100).
5. Substitute the values into the formula:
- [tex]\[
S_{120} = \frac{120}{2} \times (2 \times 200 + (120 - 1) \times 100)
\][/tex]
6. Simplify the expression:
- [tex]\[
S_{120} = 60 \times (400 + 11900)
\][/tex]
- [tex]\[
S_{120} = 60 \times 12300
\][/tex]
- [tex]\[
S_{120} = 738000
\][/tex]
Therefore, the person saves Rs. 738,000 in a decade.
