If a man saves 4 more each year than he did the year before and if he saves 20 in the first year, after how many years will his savings be more than ` 1000 altogether?

Options

A. 19 years
B. 20 years
C. 21 years
D. 18 years

Answer

To solve the problem step by step, we need to analyze the savings pattern of the man and find out after how many years his total savings will exceed 1000.

### Step 1: Understand the Savings Pattern
The man saves 20 in the first year and saves 4 more each subsequent year. This means:
- Savings in Year 1 = 20
- Savings in Year 2 = 20 + 4 = 24
- Savings in Year 3 = 24 + 4 = 28
- Savings in Year 4 = 28 + 4 = 32
- And so on...

### Step 2: Formulate the Savings for Each Year
The savings for each year can be expressed as:
- Year 1: 20
- Year 2: 20 + 4(1) = 24
- Year 3: 20 + 4(2) = 28
- Year n: 20 + 4(n - 1)

The savings in the nth year can be simplified to:
\[ \text{Savings in Year n} = 20 + 4(n - 1) = 4n + 16 \]

### Step 3: Calculate Total Savings After n Years
The total savings after n years can be calculated by summing the savings for each year:
\[ \text{Total Savings} = \text{Savings Year 1} + \text{Savings Year 2} + \ldots + \text{Savings Year n} \]
This can be expressed as:
\[ \text{Total Savings} = 20 + 24 + 28 + \ldots + (4n + 16) \]

This is an arithmetic series where:
- First term (a) = 20
- Common difference (d) = 4
- Number of terms (n)

The sum of the first n terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values:
\[ S_n = \frac{n}{2} \times (2 \times 20 + (n - 1) \times 4) \]
\[ S_n = \frac{n}{2} \times (40 + 4n - 4) \]
\[ S_n = \frac{n}{2} \times (4n + 36) \]
\[ S_n = 2n(n + 9) \]

### Step 4: Set Up the Inequality
We need to find when the total savings exceeds 1000:
\[ 2n(n + 9) > 1000 \]
Dividing both sides by 2:
\[ n(n + 9) > 500 \]

### Step 5: Solve the Quadratic Inequality
Rearranging gives us:
\[ n^2 + 9n - 500 > 0 \]
Now we can find the roots of the equation \(n^2 + 9n - 500 = 0\) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 9\), and \(c = -500\):
\[ n = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \]
\[ n = \frac{-9 \pm \sqrt{81 + 2000}}{2} \]
\[ n = \frac{-9 \pm \sqrt{2081}}{2} \]

Calculating \(\sqrt{2081} \approx 45.6\):
\[ n = \frac{-9 + 45.6}{2} \approx \frac{36.6}{2} \approx 18.3 \]
Since n must be a whole number, we round up to 19.

### Step 6: Conclusion
Thus, after 19 years, the man's total savings will exceed 1000.