The terms of a G.P. are positive. If each ter | vedclass

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

Question

(a) Under condition ${T_n} = {T_{n + 1}} + {T_{n + 2}}$

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} $

$\Rightarrow {r^{n - 1}} = {r^n}(1

Vedclass · Accepted Answer

$\frac{{\sqrt 5 - 1}}{2}$

Vedclass · Answer

(a) Under condition ${T_n} = {T_{n + 1}} + {T_{n + 2}}$

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} $

$\Rightarrow {r^{n - 1}} = {r^n}(1 + r)$

$ \Rightarrow $ ${r^2} + r - 1 = 0$

$ \Rightarrow r = \frac{{ - 1 \pm \sqrt {1 + 4} }}{2} = \frac{{ - 1 \pm \sqrt 5 }}{2}$

Since, each term is $ + ve$.

Hence common ratio is $\frac{{\sqrt 5 - 1}}{2}$.

The terms of a $G.P.$ are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is

Similar Questions

If the sum and product of four positive consecutive terms of a $G.P.$, are $126$ and $1296$, respectively, then the sum of common ratios of all such $GPs$ is $.........$.

Let the positive numbers $a _1, a _2, a _3, a _4$ and $a _5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{ m }{ n }$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$, then $m + n$ is equal to $.........$.

In a geometric progression, if the ratio of the sum of first $5$ terms to the sum of their reciprocals is $49$, and the sum of the first and the third term is $35$ . Then the first term of this geometric progression is

Insert two numbers between $3$ and $81$ so that the resulting sequence is $G.P.$

If ${x_r} = \cos (\pi /{3^r}) - i\sin (\pi /{3^r}),$ (where $i = \sqrt{-1}),$ then value of $x_1.x_2.x_3......\infty ,$ is :-