Question
Four numbers are arranged in ascending order. When the
1st two numbers are increased by the number equal to their position, and the last two numbers (i.e., 3rd and 4th numbers) are increased by 3 and 5 respectively, the series thus formed is in A.P. If the 1st number is 3 less than the 2nd number, and the sum of all the numbers is 24, then find the average of the last three numbers.Solution
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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