In a G.P.,the sum of three numbers is 14. If | vedclass

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(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.Condition $I$: $\frac{a}{r} + a + ar = 14$$\Rightarrow a(\frac{1}{r} + 1 + r) = 14$ ... $(

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$8$

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(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.Condition $I$: $\frac{a}{r} + a + ar = 14$$\Rightarrow a(\frac{1}{r} + 1 + r) = 14$ ... $(i)$Condition $II$: $\frac{a}{r} + 1, a + 1, ar - 1$ are in $A.P.$Therefore,$2(a + 1) = (\frac{a}{r} + 1) + (ar - 1)$$2a + 2 = \frac{a}{r} + ar$$2a + 2 = a(\frac{1}{r} + r)$ ... $(ii)$From $(i)$,$a(\frac{1}{r} + r) = 14 - a$. Substituting this into $(ii)$:$2a + 2 = 14 - a$$3a = 12 \Rightarrow a = 4$.Substituting $a = 4$ into $(i)$:$4(\frac{1}{r} + 1 + r) = 14$$\frac{1}{r} + 1 + r = 3.5$$r + \frac{1}{r} = 2.5$$2r^2 - 5r + 2 = 0$$(2r - 1)(r - 2) = 0$So,$r = 2$ or $r = 0.5$.If $r = 2$,the numbers are $2, 4, 8$.If $r = 0.5$,the numbers are $8, 4, 2$.In both cases,the greatest number is $8$.

In a $G.P.$,the sum of three numbers is $14$. If $1$ is added to the first two numbers and subtracted from the third number,the series becomes an $A.P.$. Then,the greatest number is:

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