Question

Don saves $1 the first week, 2$ the second week, $3 the third week, and $4 the fourth week. If he continues in this way, how many weeks will it take before Don has saved at least $50?

Please show me the simplest and quickest way to solve this problem. EXPLAIN YOUR STEPS THOROUGHLY!!

Answer 1

This forms an arithmetic sequence, where the first term = 1, common difference = 1, and number of terms = n.

The formula for the sum of an arithmetic sequence is


`S_(n) = (1)/(2)n(2a + (n - 1)d)`

Where a = the first term and d = the common difference.

We want the sum to be at least $50, so

`50 \leq (1)/(2)n(2a + (n - 1)d)`

Substituting in a and d

`50 \leq (1)/(2)n(2(1) + (n - 1)(1))`

Rearranging

`50 \leq (1)/(2)n(2 + (n - 1))`

`50 \leq (1)/(2)n(1 + n)`

`100 \leq n(1 + n)`

`100 \leq n^2 + n`

`0 \leq n^2 + n - 100`


`\text{Let } n^2 + n -100 = 0`

`n = (-1 \pm √(1-4(1)(-100)))/(2)`

`n = (-1 \pm √(401))/(2)`

`n \geq 0 \implies n = (-1 + √(401))/(2)`

`n \approx 9.51`


So it will take 9.51 weeks, or 10 weeks to the nearest week.