You have $36,948.61 in a brokerage account, and you plan to deposit an additional $3,000 at the end of every future year until your account totals $280,000. you expect to earn 11% annually on the account. how many years will it take to reach your goal? round your answer to two decimal places at the end of the calculations.

Current amount in account
P=36948.61

Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n

Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11

We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000

We can solve this by trial and error.

The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000.  It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.

So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.

So it takes 15.00 years to reach the goal of 280000 years.

The correct answer would be shape of function b = 2× slope of function a.