1. it becomes an arithmetic series
so you want:
5 + 10 + 15 + ... = 680 , a = 5 , d = 5 , n = ?
we don't know how many terms,
let n be the number of terms
(n/2)(2a + (n-1)d ) ≥ 680
(n/2)(10 + 5(n-1) ≥ 680
n(5n + 5) ≥ 1360
5n^2 + 5n - 1360 ≥ 0
n^2 + n - 272 ≥ 0
(n+17)(n-16) ≥ 0
critical values are n = -17 or n=16
but n has to be a whole number,
It will take him 16 months
2) this is a geometric series
a = 15000
r = 1.04
first year = 15000
2nd year = 15000(1.04) = ..
3rd year = 15000(1.04)^2
..
10th year = 15000(1.04)^9 = $21,349.68
sum of 10 terms
= a(r^10 - 1)/(r-1)
= 15000(1.04^10 - 1)/.04 = $180,091.61
Show the first three terms of each related sequence or series and solve each problem.
1) Jerry hopes to buy a camera which costs $680. He plans to save $5 more each month than in the month before, beginning with $5 this month. In how many months will he have saved enough?
2) A job ad promises a starting salary of $15,000 a year and a guaranteed annual raise of 4% of the previous year's salary. If you get this job, what salary should you expect for the 10th year? What would be your total earnings (before taxes) for the first 10 years?
1 answer
4 answers