A man is able to save 50 Of his salary in a particular year.After this every year he saved 20 more than the preceeding year.How long does it take him to save 4370

n/2 (50 + (n-1)*20) = 4370

Yes

To determine how long it would take for the man to save $4370, we need to find the number of years it would take for his savings to reach that amount.

Let's break down the problem step-by-step:

1. The man is able to save $50 of his salary in the first year.
2. In the second year, he saves $20 more than the previous year, so he saves $50 + $20 = $70.
3. In the third year, he saves $70 + $20 = $90.
4. We can observe that each year he saves an additional $20. So, in the fourth year, he would save $90 + $20 = $110.
5. Following this pattern, we find that his savings in subsequent years would be: $130, $150, $170, $190, and so on.

Now, our task is to find the number of years it would take for his savings to reach $4370.

Let's determine the number of years it takes to reach $4370 step-by-step:

1. Subtract $50 (initial savings) from $4370: $4370 - $50 = $4320.
2. Divide the remaining amount by $20 to find the number of increments of $20: $4320 / $20 = 216.
3. Add 1 to account for the initial year: 216 + 1 = 217.

Therefore, it would take the man 217 years to save $4370.

To determine how long it takes for the man to save $4,370, we need to break down the problem into smaller parts and solve it step by step.

Let's consider the first year of saving. It's given that the man is able to save 50% of his salary. Let's assume his salary in the first year is S1. Therefore, he saves 50% of S1, which is 0.5S1.

In the second year, the man saves 20 more than the previous year. So in the second year, he saves 0.5S1 + 20.

Similarly, in the third year, he saves 0.5S1 + 20 + 20, which simplifies to 0.5S1 + 40.

Using this pattern, we can find the savings for subsequent years. In general, the savings for the nth year can be calculated as:
Savings for year n = 0.5S1 + 20(n-1)

Now, we need to find the number of years it takes for the man's total savings to reach $4,370. To do this, we set up the following equation:

0.5S1 + 20(1-1) + 0.5S1 + 20(2-1) + 0.5S1 + 20(3-1) + ... + 0.5S1 + 20(n-1) = 4,370

To simplify this equation, notice that the terms 20(1-1), 20(2-1), 20(3-1), etc., can be replaced by a simpler expression, 20(0), 20(1), 20(2), etc. This realization allows us to simplify the equation:

0.5S1 + 20(0) + 0.5S1 + 20(1) + 0.5S1 + 20(2) + ... + 0.5S1 + 20(n-1) = 4,370

0.5S1 + 0.5S1 + 0.5S1 + ... + 0.5S1 + 20(0 + 1 + 2 + ... + (n-1)) = 4,370

0.5S1n + 20(0 + 1 + 2 + ... + (n-1)) = 4,370

0.5S1n + 10(n-1)(n) = 4,370

Simplifying further, we get:

0.5S1n + 5(n^2 - n) = 4,370

0.5S1n + 5n^2 - 5n = 4,370

Rearranging the terms, we have:

5n^2 - 4.5n - 4,370 = 0

To solve this quadratic equation, we can use the quadratic formula:

n = (-b ± sqrt(b^2 - 4ac)) / (2a)

Where a = 5, b = -4.5, and c = -4,370

Plugging in the values, we find:

n = (-(-4.5) ± sqrt((-4.5)^2 - 4(5)(-4,370))) / (2(5))

Solving this equation, we get two possible values for n: n ≈ 20.1 and n ≈ -86.1.

Since time cannot be negative, we discard the negative value. Therefore, it takes approximately 20 years for the man to save $4,370.