The sum of three consecutive terms in a geome | vedclass

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(B) Let the three consecutive terms of the geometric progression be $a/r, a, ar$.Given that their sum is $14$,so $a/r + a + ar = 14$,which implies $a(

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$2$

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(B) Let the three consecutive terms of the geometric progression be $a/r, a, ar$.Given that their sum is $14$,so $a/r + a + ar = 14$,which implies $a(1/r + 1 + r) = 14$ ..... $(i)$According to the problem,if $1$ is added to the first and second terms and $1$ is subtracted from the third,the new terms are $a/r + 1, a + 1, ar - 1$.Since these are in arithmetic progression,$2(a + 1) = (a/r + 1) + (ar - 1)$.$2a + 2 = a/r + ar$.$2a + 2 = a(1/r + r)$.From $(i)$,$a(1/r + r) = 14 - a$. Substituting this into the equation:$2a + 2 = 14 - a$.$3a = 12$,so $a = 4$.Substituting $a = 4$ into $(i)$: $4(1/r + 1 + r) = 14$.$1/r + 1 + r = 3.5$.$1/r + r = 2.5$.$r^2 - 2.5r + 1 = 0$.$2r^2 - 5r + 2 = 0$.$(2r - 1)(r - 2) = 0$.So,$r = 2$ or $r = 1/2$.If $r = 2$,the terms are $4/2, 4, 4(2)$,which are $2, 4, 8$.If $r = 1/2$,the terms are $4/(1/2), 4, 4(1/2)$,which are $8, 4, 2$.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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