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Question
The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
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Solution
Let the first term of an A.P is a and its common difference be d.
\[\therefore a_1 + a_2 + a_3 = 21\]
\[ \Rightarrow a + \left( a + d \right) + \left( a + 2d \right) = 21\]
\[ \Rightarrow 3a + 3d = 21 \]
\[ \Rightarrow a + d = 7 . . . (i)\]
\[\text { Now, according to the question }: \]
\[a , a + d - 1 \text { and } a + 2d + 1 \text { are in G . P } . \]
\[ \Rightarrow \left( a + d - 1 \right)^2 = a\left( a + 2d + 1 \right)\]
\[ \Rightarrow \left( 7 + a - a - 1 \right)^2 = a \left[ a + 2\left( 7 - a \right) + 1 \right] \]
\[ \Rightarrow \left( 6 \right)^2 = a\left( 15 - a \right)\]
\[ \Rightarrow 36 = 15a - a^2 \]
\[ \Rightarrow a^2 - 15a + 36 = 0\]
\[ \Rightarrow \left( a - 3 \right)\left( a - 12 \right) = 0\]
\[ \Rightarrow a = 3, 12\]
\[\text { Now, putting a = 2, 12 in equation (i), we get d = 5, - 5, respectively } . \]
\[\text { Thus, for a = 2 and d = 5, the numbers are 2, 7 and 12 } . \]
\[\text { And, for a = 12 and d = - 5, the numbers are 12 , 7 and 2 } . \]
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