The sum of three consecutive terms in a geome | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

If for $x, y \in \mathbb{R}, x > 0,$ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ up to $\infty$ terms and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$,then the ordered pair $(x, y)$ is equal to:

Let $3, 6, 9, 12, \ldots$ up to $78$ terms and $5, 9, 13, 17, \ldots$ up to $59$ terms be two series. Then,the sum of the terms common to both the series is equal to

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

If $a, b, c$ are in $GP$ and $4a, 5b, 4c$ are in $AP$ such that $a + b + c = 70$,then the value of $a^3 + b^3 + c^3$ is

If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$,$y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$,$z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta$ and $0 < \theta < \frac{\pi}{2}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module