Find the 7th term of G.P. from the following sequence: 4, 8, 16, 32. If 22x+y = 32 and 2^x−y = 1, find the value of x and y.

The first term is $a = 4$.
The common ratio $r$ is found by dividing any term by its preceding term:

$$r = \frac{8}{4} = 2$$

We need to find the 7th term, so $n=7$.

$$T_7 = 4 \cdot 2^{7-1}$$ $$T_7 = 4 \cdot 2^6$$

$$T_7 = 4 \cdot 64$$ $$T_7 = 256$$

The 7th term is 256.

The correct option is (C).
The final answer is $\boxed{\text{C}}$

For the first equation:

$$2^{2x+y} = 32$$

Since $32 = 2^5$, we have:

$$2^{2x+y} = 2^5$$

Equating the exponents gives:

$$2x+y = 5 \quad (\text{Equation 1})$$

For the second equation:

$$2^{x-y} = 1$$

Since $1 = 2^0$, we have:

$$2^{x-y} = 2^0$$

Equating the exponents gives:

$$x-y = 0 \quad (\text{Equation 2})$$

From Equation 2, we can deduce:

$$x = y$$

Substitute $x$ for $y$ in Equation 1:

$$2x + x = 5$$ $$3x = 5$$ $$x = \frac{5}{3}$$

Since $x=y$, then:

$$y = \frac{5}{3}$$

The values are $x = \frac{5}{3}$ and $y = \frac{5}{3}$.
Upon reviewing the provided options, none of them exactly match the calculated values of $x = \frac{5}{3}$ and $y = \frac{5}{3}$.
The final answer is $\boxed{x = \frac{5}{3}, y = \frac{5}{3}}$