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Questions
The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.
The sum of three consecutive numbers of a G.P. is 14. If 1 is added in first and second term each and 1 subtracted from third, the new numbers form an A.P. Find the numbers.
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Solution
Let the numbers be a, ar and ar2.
\[\text { Sum }= 14 \]
\[ \Rightarrow a + ar + a r^2 = 14 \]
\[ \Rightarrow a(1 + r + r^2 ) = 14 . . . \left( i \right)\]
According to the question,a + 1, ar + 1 and ar2 − 1 are in A.P.
\[\therefore 2\left( ar + 1 \right) = a + 1 + a r^2 - 1\]
\[ \Rightarrow 2ar + 2 = a + a r^2 \]
\[ \Rightarrow 2ar + 2 = 14 - ar [\text { From }\left( i \right)]\]
\[ \Rightarrow 3ar = 12 \]
\[ \Rightarrow a = \frac{4}{r} . . . \left( ii \right)\]
\[\text { Putting } a = \frac{4}{r} \text { in }\left( i \right)\]
\[ \Rightarrow \frac{4}{r}(1 + r + r^2 ) = 14\]
\[ \Rightarrow 4 r^2 - 10r + 4 = 0 \]
\[ \Rightarrow 4 r^2 - 8r - 2r + 4 = 0 \]
\[ \Rightarrow \left( 4r - 2 \right)\left( r - 2 \right) = 0\]
\[ \Rightarrow r = \frac{1}{2}, 2\]
\[\text { Putting } r = \frac{1}{2}\text { in } \left( ii \right), \text { we get a } = 8 . \]
\[\text { So, the G . P . is 8, 4 and } 2 . \]
Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.
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