To solve for \( a_1 \) given the recursive equation \( a_n = a_{n-1} + 7 \) for \( n \geq 1 \) and the condition \( a_5 = 43 \), we can work backwards through the sequence.
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Starting from \( a_5 \): \[ a_5 = a_4 + 7 \] Since \( a_5 = 43 \): \[ 43 = a_4 + 7 \implies a_4 = 43 - 7 = 36 \]
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Now calculate \( a_4 \): \[ a_4 = a_3 + 7 \] From above, \( a_4 = 36 \): \[ 36 = a_3 + 7 \implies a_3 = 36 - 7 = 29 \]
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Calculate \( a_3 \): \[ a_3 = a_2 + 7 \] From above, \( a_3 = 29 \): \[ 29 = a_2 + 7 \implies a_2 = 29 - 7 = 22 \]
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Calculate \( a_2 \): \[ a_2 = a_1 + 7 \] From above, \( a_2 = 22 \): \[ 22 = a_1 + 7 \implies a_1 = 22 - 7 = 15 \]
Thus, we find that \( a_1 = 15 \).
The answer is \( \boxed{15} \).