The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

#### Solution

Let the first term of the A.P. be a and the common difference be d.

∴ a = a , b = a + d and c = a + 2d

\[a + b + c = 18\]

\[ \Rightarrow a + \left( a + d \right) + \left( a + 2d \right) = 18\]

\[ \Rightarrow 3a + 3d = 18 \]

\[ \Rightarrow a + d = 6 . . . . . . . (i)\]

\[\text { Now, according to the question, a + 4, a + d + 4 and a + 2d + 36 are in G . P .} \]

\[ \therefore \left( a + d + 4 \right)^2 = \left( a + 4 \right)\left( a + 2d + 36 \right)\]

\[ \Rightarrow \left( 6 - d + d + 4 \right)^2 = \left( 6 - d + 4 \right) \left( 6 - d + 2d + 36 \right) \]

\[ \Rightarrow \left( 10 \right)^2 = \left( 10 - d \right)\left( 42 + d \right)\]

\[ \Rightarrow 100 = 420 + 10d - 42d - d^2 \]

\[ \Rightarrow d^2 + 32d - 320 = 0\]

\[ \Rightarrow \left( d + 40 \right)\left( d - 8 \right) = 0\]

\[ \Rightarrow d = 8, - 40\]

\[\text { Now, putting d = 8, - 40 in equation (i), we get, a = - 2, 46, respectively .} \]

\[\text { For a = - 2 and d = 8, we have }: \]

\[ a = - 2 , b = 6 , c = 14\]

\[\text { And, for a = 46 and d = - 40, we have }: \]

\[ a = 46 , b = 6 , c = - 34\]