Suppose that $a(t) = 0.1t^2 + 1$. At time 0, 1000 is invested. An additional investment of X is made at time 6. If the total accumulated value of these two investments at time t=8 is 18800, what is X?
My work: From original investment, at time t=6, it is valued at$A(6) = 1000a(6)= 4600$. Thus, we solve for $A(8) = 18800 = (4600+X) \frac{a(8)}{a(6)}?$
The correct answer is 6589.
Yes, you are right. But only if " the total accumulated value of these two investments at time $t=8$ is $\color{red}{18000}$". The equation is
$$(4600+X)\cdot (0.1\cdot 8^2+1)/(0.1\cdot 6^2+1)=18000$$
$$4600+X=18000\cdot (0.1\cdot 6^2+1)/(0.1\cdot 8^2+1)$$
$$X=18000\cdot (0.1\cdot 6^2+1)/(0.1\cdot 8^2+1)-4600=6589.\overline{189}\approx 6589$$