 Solution

Adjacent slit contributions have phase difference $\delta =2\pi d\sin \theta /\lambda$. The six-slit amplitude is a geometric sum ${\sum }_{j=0}^5{e}^{ij\delta }=e^{5i\delta /2}\sin (3\delta )/\sin (\delta /2)$. The first zero away from the central maximum occurs at $\delta =2\pi /6$, where the denominator is nonzero. ThereforeThere is a matching zero at the negative angle. This is the first minimum of a finite slit array, rather than the much larger principal-order angle or a single-slit envelope minimum. The slits are stipulated sufficiently narrow for that envelope not to set the first zero.