Verified Answer :

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\(\left \{ {{a_1=x} \atop {a_{n}=2a_{n-1}-3}} \right. \\ \\ a_2=2a_1-3=2x-3 \\ a_3=2a_2-3=2(2x-3)-3=4x-6-3=4x-9 \\ a_4=2a_3-3=2(4x-9)-3=8x-18-3=8x-21 \\ a_5=2a_4-3=2(8x-21)-3=16x-42-3=16x-45 \\ \\ \boxed{a_5=16x-45}\)

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