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Present Worth of an Arithmetic Gradient

Hardarithmetic gradientpresent worth
Maintenance on a bridge costs $2,000 in year 1 and increases by $500 each year through year 10. At 8%, the present worth of the maintenance costs is most nearly:
  1. A$13,420
  2. B$28,518
  3. C$26,409
  4. D$42,500
Show solution

Answer: C. $26,409

Equation: P = A(P/A, i, n) + G(P/G, i, n). Values: base amount A = $2,000 in year 1, gradient G = $500 per year, i = 8%, n = 10; (P/A, 8%, 10) = 6.7101 and (P/G, 8%, 10) = 25.9768. Step 1: 2,000 × 6.7101 = $13,420. Step 2: 500 × 25.9768 = $12,988. Step 3: P = 13,420 + 12,988 = $26,409. $13,420 ignores the gradient; $42,500 adds the payments with no discounting; $28,518 treats the gradient as a uniform series.
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