Consider initializing two pointers: one at the start (`left`) and one at the end (`right`) of the array.

Calculate the sum of the elements at these pointers. If the sum equals the target, return their 1-based indices. If the sum is less than the target, move the `left` pointer right to increase the sum. If the sum is greater, move the `right` pointer left to decrease the sum.

Since the array is sorted and exactly one solution exists, the loop will terminate when the correct pair is found, ensuring optimal O(n) time complexity with O(1) space.