Question

Difficulty: HardArithmetic and Geometric Sequences and Series

An entrepreneur starts a company with an operating budget of 100,000100,000 in its first year. For each of the next 4 years (years 2 through 5), the budget increases by a constant amount of dd dollars each year. For years 6 through 8, the budget increases geometrically, where the budget in year 6 is 1.51.5 times the budget in year 5, and the budget increases by 50%50\% each year thereafter. If the total operating budget over the first 8 years is 1,597,5001,597,500 dollars, what is the value of dd?

  1. A
    4,3004,300
  2. 10,00010,000Answer
  3. C
    17,50017,500
  4. D
    33,90033,900
  5. E
    38,50038,500

Answer

The common difference is 10,00010,000.
The correct answer is found by setting up the sum of the first 5 years of the arithmetic sequence and the subsequent 3 years of the geometric sequence. Summing the expressions for all 8 years gives 1,212,500+38.5d1,212,500 + 38.5d. Equating this expression to the total budget of 1,597,5001,597,500 and solving for dd yields 10,00010,000.

Step-by-Step Solution

1
Express the budget for the first 5 years as an arithmetic sequence and find their sum.
S5=100,000+(100,000+d)+(100,000+2d)+(100,000+3d)+(100,000+4d)=500,000+10dS_5 = 100,000 + (100,000 + d) + (100,000 + 2d) + (100,000 + 3d) + (100,000 + 4d) = 500,000 + 10d
The budget starts at 100,000100,000 in year 1 and increases by a constant amount dd each year through year 5.
2
Express the budgets for years 6 through 8 as a geometric sequence starting from 1.51.5 times the year 5 budget.
Year 6: 1.5(100,000+4d)=150,000+6d1.5(100,000 + 4d) = 150,000 + 6d; Year 7: 1.5(150,000+6d)=225,000+9d1.5(150,000 + 6d) = 225,000 + 9d; Year 8: 1.5(225,000+9d)=337,500+13.5d1.5(225,000 + 9d) = 337,500 + 13.5d. The sum of these 3 years is 712,500+28.5d712,500 + 28.5d.
The budget increases by a factor of 1.51.5 (or 50%50\%) each year starting from year 6.
3
Combine the sums of both sequences to represent the total 8-year budget and solve for dd.
Total = (500,000+10d)+(712,500+28.5d)=1,212,500+38.5d(500,000 + 10d) + (712,500 + 28.5d) = 1,212,500 + 38.5d. Setting this equal to the given total: 1,212,500+38.5d=1,597,500    38.5d=385,000    d=10,0001,212,500 + 38.5d = 1,597,500 \implies 38.5d = 385,000 \implies d = 10,000.
The total budget over the 8 years is the sum of the budgets of the individual years.

Key Concept

Combining arithmetic and geometric sequences in multi-step word problems.

Alternative Method

Instead of calculating each geometric year sequentially, the sum of the geometric sequence for years 6 to 8 can be calculated using the geometric series sum formula Sn=a11rn1rS_n = a_1 \frac{1 - r^n}{1 - r} with a1=1.5(100,000+4d)a_1 = 1.5(100,000 + 4d) and r=1.5r = 1.5 over n=3n = 3 terms.
Estimated Time:2m 30s
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