A student deposits the same amount of money into her bank account each week. At the end of the second week she has $30 in her account. At the end of the third week she has $45 in her account. How much will she have in her bank account at the end of the ninth week?

Question

bloomlocke367 · Answer

Should I just write an explicit formula?

igreen · Answer

Week 2 = 30
Week 3 = 45

I guess we can make an equation..

(2, 30) and (3, 45)

Plug them into the slope formula:

$m = \dfrac{y_2-y_1}{x_2-x_1}$

$m = \dfrac{45-30}{3-2}$

Simplify

bloomlocke367 · Answer

I'm assuming that at the end of the first week she had $15... should I make that assumption? and why did you make that equation? Would it be wrong to use:$$a _{n}=a _{1}+(n-1)d$$

igreen · Answer

No..it wouldn't.

bloomlocke367 · Answer

Okay. Which is easier?

igreen · Answer

Well, just use the formula since that's what they want you to do.

igreen · Answer

'a' is the first term, which you are correct, is 15.

And 'd' is the common difference.

igreen · Answer

To find 'd', just subtract 45 - 30

bloomlocke367 · Answer

$$a _{n}=15+(n-1)15$$
$$ =15+15n-15$$
$$=15n$$
right?

igreen · Answer

Yep, now plug in '9' for 'n'

igreen · Answer

That will give us the amount she has on the 9th week.

bloomlocke367 · Answer

135

igreen · Answer

Correct

bloomlocke367 · Answer

That was easier than I thought... I saw the word problem and kinda freaked out XD

igreen · Answer

Finding the slope would've gave us 15..

$m = \dfrac{y_2-y_1}{x_2-x_1}$

$m = \dfrac{45-30}{3-2}$

$m = \dfrac{15}{1}$

$m = 15$

So the equation would've been y = 15x, and we plug in '9' for 'x' and we get the same thing :P

bloomlocke367 · Answer

Okay. :) thanks.