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Question
Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.
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Solution
Let the first term of an A.P. be a and its common difference be d.
\[a_1 + a_2 + a_3 = 15\]
\[ \Rightarrow a + \left( a + d \right) + \left( a + 2d \right) = 15\]
\[ \Rightarrow 3a + 3d = 15 \]
\[ \Rightarrow a + d = 5 . . . . . . . (i)\]
\[\text { Now, according to the question }: \]
\[a + 1, a + d + 3 \text { and }a + 2d + 9 \text { are in G . P } . \]
\[ \Rightarrow \left( a + d + 3 \right)^2 = \left( a + 1 \right)\left( a + 2d + 9 \right)\]
\[ \Rightarrow \left( 5 - d + d + 3 \right)^2 = \left( 5 - d + 1 \right) \left( 5 - d + 2d + 9 \right) \left[ \text { From } (i) \right] \]
\[ \Rightarrow \left( 8 \right)^2 = \left( 6 - d \right)\left( 14 + d \right)\]
\[ \Rightarrow 64 = 84 + 6d - 14d - d^2 \]
\[ \Rightarrow d^2 + 8d - 20 = 0\]
\[ \Rightarrow \left( d - 2 \right)\left( d + 10 \right) = 0\]
\[ \Rightarrow d = 2, - 10\]
\[\text { Now, putting } d = 2, - 10 \text { in equation (i), we get, a } = 3, 15,\text { respectively } . \]
\[\text { Thus, for } a = 3 \text { and }d = 2, \text { the A . P . is } 3, 5, 7 . \]
\[\text { And, for a = 15 and d = - 10, the A . P . is }15 , 5, - 5 . \]
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