The sum of three consecutive terms in a geome | vedclass

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(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.Given the sum is $14$,so $\frac{a}{r} + a + ar = 14 \Rightarrow a(\frac{1}{r}

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$2$

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(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.Given the sum is $14$,so $\frac{a}{r} + a + ar = 14 \Rightarrow a(\frac{1}{r} + 1 + r) = 14$ ... $(i)$According to the problem,the new terms are $(\frac{a}{r} + 1), (a + 1), (ar - 1)$,which are in arithmetic progression.Therefore,$2(a + 1) = (\frac{a}{r} + 1) + (ar - 1)$$2a + 2 = \frac{a}{r} + ar$$2a + 2 = a(\frac{1}{r} + r)$From $(i)$,$a(\frac{1}{r} + r) = 14 - a$. Substituting this into the equation:$2a + 2 = 14 - a$$3a = 12 \Rightarrow a = 4$.Now,substitute $a = 4$ into $(i)$:$4(\frac{1}{r} + 1 + r) = 14$$\frac{1}{r} + 1 + r = 3.5$$r + \frac{1}{r} = 2.5 = \frac{5}{2}$$2r^2 - 5r + 2 = 0$$(2r - 1)(r - 2) = 0$So,$r = 2$ or $r = \frac{1}{2}$.If $r = 2$,the terms are $\frac{4}{2}, 4, 4(2) = 2, 4, 8$.If $r = \frac{1}{2}$,the terms are $\frac{4}{1/2}, 4, 4(1/2) = 8, 4, 2$.In both cases,the terms are $2, 4, 8$. The lowest term is $2$.

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

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