## Algebra and Trigonometry 10th Edition

$A_{80}\ne 2\times A_{40}$
$A_n=10,000(1+\frac{0.035}{4})^n$ In order to find the balance after 20 years we need to compute the 80th term of the sequence: $A_{80}=10,000(1+\frac{0.035}{4})^{80}=10,000(1.00875)^{80}=20076.31$ From item (b): $A_{40}=14,169.09$ It is clear that $A_{80}\ne 2\times A_{40}=28,338.18$