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Let the four numbers be a,b,c,d. Given: a=b−3, a+b+c+d=24, and new series (a+1,b+2,c+3,d+5) forms an A.P. A.P. Condition:(b+2) − (a+1)=4, (c+3) − (b+2)=4, (d+5) − (c+3) = 4. This gives: c − b = 3 and d−c=2. Substitute c=b+3 and d=b+5 into the sum equation: a+b+c+d=24 ⟹ (b−3)+b+(b+3)+(b+5)=24. Solve: 4b+5=24 ⟹ b = 4.75. Find other numbers: a = b − 3 = 1.75, c = b+3 = 7.75,d = b+5 = 9.75. Average of the last three numbers: 
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