A one-dimensional array A has indices 1 through 75. Each element is a string…

Concept

For a one-dimensional array with lower bound L and element width w, the address of A[i] is found by adding the storage occupied by the preceding elements to the base address.

Therefore, Address(A[i]) = Base + (i − L) × w. The subtraction counts how many complete elements come before A[i].

Application

1. Identify the values: Base = 1120, i = 49, L = 1, and w = 3 memory words.

2. Count the preceding elements: i − L = 49 − 1 = 48.

3. Find their storage: 48 × 3 = 144 memory words.

4. Add the offset to the base: 1120 + 144 = 1264.

Cross-check

A[1] starts at 1120. Moving forward by 48 elements, with 3 words per element, advances the address by 144, so A[49] starts at 1264.