The sum of the first ten terms of an A.P. is 160 and the sum of the...

Let the first term of the A.P. be $a$ and its common difference be $d$.
Let the first term of the G.P. be $A$ and its common ratio be $R$.

Given that the sum of the first ten terms of the A.P. is $160$:
$\dfrac{10}{2} [2a + (10 - 1)d] = 160$
$5(2a + 9d) = 160$
$2a + 9d = 32$

Given that the sum of the first two terms of the G.P. is $8$:
$A + AR = 8$
$A(1 + R) = 8$

We are also given that $a = R$ and $A = d$. Substituting these into the first equation:
$2R + 9A = 32$
$R = \dfrac{32 - 9A}{2}$

Substitute $R$ into the second equation:
$A \left(1 + \dfrac{32 - 9A}{2}\right) = 8$
$A \left(\dfrac{2 + 32 - 9A}{2}\right) = 8$
$A(34 - 9A) = 16$
$34A - 9A^2 = 16$
$9A^2 - 34A + 16 = 0$

This is a quadratic equation in $A$. The discriminant is $\Delta = (-34)^2 - 4(9)(16) = 1156 - 576 = 580 > 0$, so there are two distinct real values for $A$.

The sum of all possible values of the first term of the G.P. ($A$) is the sum of the roots of this quadratic equation:
Sum of roots $= -\dfrac{\text{coefficient of } A}{\text{coefficient of } A^2} = \dfrac{34}{9}$

Answer: $\dfrac{34}{9}$