= Solution
On a finite horizon $[0,T]$, let $W$ be a standard $d$-dimensional <Brownian motion> and let $\theta$ be a <predictable process> with $\int_0^T|\theta_s|^2ds<\infty$ almost surely. Suppose the <stochastic exponential>
$$
Z_t=\exp\left(-\int_0^t\theta_s\cdot dW_s-\frac12\int_0^t|\theta_s|^2ds\right)
$$
is a true <martingale> with $\mathbb EZ_T=1$. The <Novikov condition> $\mathbb E\exp(\frac12\int_0^T|\theta_s|^2ds)<\infty$ is a sufficient condition. The <Girsanov theorem> says that under $dQ=Z_TdP$,
$$
\boxed{W_t^Q=W_t+\int_0^t\theta_sds\text{ is standard Brownian motion}.}
$$
Because $Z_T>0$, the two measures are equivalent on the terminal sigma-algebra.
For the proof sketch, <Itô formula> gives $dZ_t=-Z_t\theta_t\cdot dW_t$. Apply the product rule to $Z_tW_t^{Q,i}$. Its finite-variation term $Z_t\theta_t^i dt$ is cancelled by $d[Z,W^i]_t=-Z_t\theta_t^i dt$. After localization the product is a <martingale> under $P$, and the conditional change-of-measure formula makes $W^{Q,i}$ a <local martingale> under $Q$. Adding finite variation leaves its covariations unchanged: $[W^{Q,i},W^{Q,j}]_t=\delta_{ij}t$. The <Lévy characterization of multidimensional Brownian motion> completes the proof. Localization is essential if the integrands are not bounded; the true-martingale hypothesis on $Z$ permits the final measure change.
For the deal, write $T=t_0$ and use constant <interest rate> $\rho$, volatility $\sigma>0$ and no <stock> dividends. Under the <risk-neutral measure>, the <Black-Scholes model> is
$$
S_t=S_0\exp\left((\rho-\tfrac12\sigma^2)t+\sigma W_t^Q\right).
$$
For example this follows from the preceding theorem with $\theta=(\mu-\rho)/\sigma$ when the physical <stock> drift is $\mu$. The delivered shares have terminal cash value $S_T\max(r,c/A)$, so its initial value is
$$
V_0=e^{-\rho T}\mathbb E_Q\left[S_T\max(r,c/A)\right].
$$
The factor $S_T$ must be included: the deal is a <stock> delivery, not a cash payment of the random number of shares.
Use the <stock-numeraire measure in the Black-Scholes model>, whose density is
$$
\frac{dQ^S}{dQ}=\frac{e^{-\rho T}S_T}{S_0}=\exp\left(\sigma W_T^Q-\tfrac12\sigma^2T\right).
$$
The <Girsanov theorem> gives $W_t^S=W_t^Q-\sigma t$ as <Brownian motion> under $Q^S$. Therefore
$$
\log A=\log S_0+(\rho+\tfrac12\sigma^2)\bar t+\frac\sigma n\sum_{i=1}^nW_{t_i}^S,\qquad\bar t=\frac1n\sum_{i=1}^nt_i.
$$
The <discrete geometric average under the stock-numeraire measure> has normally distributed logarithm with
$$
m=\log S_0+(\rho+\tfrac12\sigma^2)\bar t,\qquad v=\frac{\sigma^2}{n^2}\sum_{i,j=1}^n\min(t_i,t_j)=\frac{\sigma^2}{n^2}\sum_{i=1}^n(2i-1)t_i.
$$
The last identity uses the printed descending order of sampling times. Each $t_i$ is the minimum in exactly $2i-1$ ordered pairs. In particular $v>0$.
Assume the ordinary positive deal parameters $r,c>0$ and put $k=\log(c/r)$, $d=(m-k)/\sqrt v$. The <change of numeraire> gives $V_0=S_0\mathbb E_{Q^S}\max(r,ce^{-Y})$, where $Y=\log A$. Split the integral at $k$. Completing the square in $e^{-y}$ times the <normal probability density> gives
$$
\mathbb P(Y\ge k)=\Phi(d),\qquad\mathbb E[e^{-Y}\mathbf1_{Y<k}]=e^{-m+v/2}\Phi\left(\frac{k-m+v}{\sqrt v}\right).
$$
Thus the <stock-delivery option with a geometric average> has value
$$
\boxed{V_0=S_0\left[r\Phi(d)+ce^{-m+v/2}\Phi(-d+\sqrt v)\right].}
$$
Here $\Phi$ is the standard <normal distribution> function. Delivery time $T$ does not enter separately once the sampling dates are fixed: the <stock> numeraire cancels the final <stock> factor and discounting. As a check, for one sample at $T$, the payment is $\max(rS_T,c)$ and the formula becomes the value of $r$ shares plus a put on their terminal value. If $c=0$ the value is $rS_0$; if $r=0$ the value is $cS_0e^{-m+v/2}$.
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