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. 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
1) In this problem, we have a sequence where each term represents the amount of money Jerry saves each month. We are given that he plans to save $5 more each month than in the month before, beginning with $5 this month.
To find the first three terms of the sequence, we can start by listing them out:
Term 1: $5
Term 2: $5 + $5 = $10
Term 3: $10 + $5 = $15
Therefore, the first three terms of the sequence are $5, $10, and $15.
Now, we need to determine in how many months Jerry will have saved enough to buy the $680 camera. We can set up an equation to solve this:
$5 + $10 + $15 + ... = $680
To find the number of terms needed, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
where:
Sn is the sum of the first n terms,
a is the first term,
d is the common difference between terms,
and n is the number of terms.
In this case, a = $5, d = $5, and Sn = $680. Plugging in these values, we have:
680 = (n/2)(2*5 + (n-1)*5)
Simplifying the equation gives:
680 = (n/2)(10 + 5n - 5)
Now, we can solve for n by rearranging the equation:
680 = (n/2)(5n + 5)
1360 = 5n^2 + 5n
Rearranging again:
5n^2 + 5n - 1360 = 0
Now, we can use the quadratic formula to solve for n:
n = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 5, b = 5, and c = -1360. Plugging in these values, we get:
n = (-5 ± √(5^2 - 4*5*(-1360)))/(2*5)
Simplifying further gives:
n ≈ (-5 ± √(25 + 27200))/10
n ≈ (-5 ± √(27225))/10
n ≈ (-5 ± 165)/10
Now, we have two possible values for n:
n ≈ (-5 + 165)/10 ≈ 16
n ≈ (-5 - 165)/10 ≈ -17
Since the number of months cannot be negative, we discard the -17.
Therefore, Jerry will have saved enough to buy the camera in approximately 16 months.
2) In this problem, we have a series of salaries over 10 years, with a guaranteed annual raise of 4% of the previous year's salary. The starting salary is $15,000.
To find the salary for the 10th year, we can use the formula for the nth term of a geometric series:
an = a * r^(n-1)
where:
an is the nth term,
a is the first term,
r is the common ratio, which is the raise rate as a decimal,
and n is the number of terms.
In this case, a = $15,000, r = 1 + 4% = 1 + 0.04 = 1.04, and n = 10. Plugging in these values, we have:
a10 = $15,000 * (1.04)^(10-1)
Simplifying gives:
a10 = $15,000 * (1.04)^9
Calculating this value will give us the salary for the 10th year.
To find the total earnings for the first 10 years, we need to calculate the sum of the first 10 terms of the geometric series. We can use the formula for the sum of a geometric series:
Sn = a * (1 - r^n) / (1 - r)
where:
Sn is the sum of the first n terms,
a is the first term,
r is the common ratio, which is the raise rate as a decimal,
and n is the number of terms.
In this case, a = $15,000, r = 1.04, and n = 10. Plugging in these values, we have:
S10 = $15,000 * (1 - (1.04)^10) / (1 - 1.04)
Calculating this value will give us the total earnings for the first 10 years (before taxes).