Problem: Sum of three numbers in G.P be 14. If one is added to first and second and 1 is subtracted from the third, the new numbers are in A.P. The smallest of them is?

Solution:

Step 1: Find the three numbers in G.P
Let the three numbers in G.P be a, ar and ar². Therefore,
a + ar + ar² = 14

Step 2: Find the three numbers in A.P
If one is added to the first and second terms and 1 is subtracted from the third term, we get the new A.P. The three terms are a+1, ar+1 and ar²-1.
The common difference of the new A.P is (ar+1)-(a+1) = (ar-a).
Therefore,
(ar²-1)-(ar+1) = (ar+1)-(a+1)

Simplifying the above equation, we get
ar²-2ar-a = 0
Solving this quadratic equation, we get ar = (-(-2) ± √((-2)²-4(1)(-a)))/(2(1)) = (2 ± √(4+4a))/2 = 1 ± √(1+a)

Step 3: Find the three numbers in G.P using the value of ar
From Step 1, we know that a + ar + ar² = 14
Substituting ar = 1+√(1+a), we get a + (1+√(1+a)) + (1+√(1+a))² = 14
Expanding the RHS, we get a + 2(1+√(1+a)) + (1+a+2√(1+a)) = 14
Simplifying the above equation, we get a = 3-3√(1+a)

Solving the above equation, we get a = 1 or a = 4.

Therefore, the three numbers in G.P are either 1, 2+√2, 5-√2 or 4, 2-√2, 2+√2.

Step 4: Find the three numbers in A.P using the values of a and ar
Using the value of a and ar, we get the three numbers in A.P as follows:
If a = 1 and ar = 2+√2, then the three numbers in A.P are 2, 3+√2, 4+2√2.
If a = 4 and ar = 2-√2, then the three numbers in A.P are 5, 3-√2, 1-2√2.

Step 5: Find the smallest number in the A.P
The common difference of the A.P is (ar-a) = 1±√(1+a) - a = 1±√(1+a) - (1±√(1+a))²/a
Substituting the values of a and ar, we get the two common differences as √2-1 and 1-√2.

Therefore

8 or 2 for r=1/2; r=2