# S is a geometric sequence? a)Given that (sqrtx-1), 1 and (sqrtx+1) are the 1st 3 terms of S, find the value of x. b)Show that the 5th term of S is 7+5sqrt2

May 26, 2018

a)$x = 2$
b) see below

#### Explanation:

a) Since the first three terms are $\sqrt{x} - 1$, 1 and $\sqrt{x} + 1$, the middle term, 1, must be the geometric mean of the other two. Hence

${1}^{2} = \left(\sqrt{x} - 1\right) \left(\sqrt{x} + 1\right) \implies$

$1 = x - 1 \implies x = 2$

b)

The common ratio is then $\sqrt{2} + 1$, and the first term is $\sqrt{2} - 1$.

Thus, the fifth term is

$\left(\sqrt{2} - 1\right) \times {\left(\sqrt{2} + 1\right)}^{4} = {\left(\sqrt{2} + 1\right)}^{3}$
$q \quad = {\left(\sqrt{2}\right)}^{3} + 3 {\left(\sqrt{2}\right)}^{2} + 3 \left(\sqrt{2}\right) + 1$
$q \quad = 2 \sqrt{2} + 6 + 3 \sqrt{2} + 1$
$q \quad = 7 + 5 \sqrt{2}$

May 26, 2018

#### Explanation:

Given that,

$\rightarrow \sqrt{x} - 1 , 1 , \sqrt{x} + 1$ are in $G P$.

So,

$\rightarrow \frac{\sqrt{x} - 1}{1} = \frac{1}{\sqrt{x} + 1}$

$\rightarrow {\left(\sqrt{x} - 1\right)}^{2} = 1$

$\rightarrow {\left(\sqrt{x}\right)}^{2} - {1}^{2} = 1$

$\rightarrow x = 2$

The first term $\left(a\right) = \sqrt{x} - 1 = \sqrt{2} - 1$

The second term $\left(b\right) = 1$

The common ratio $\left(r\right) = \frac{b}{a} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$

The ${n}^{t h}$ term of geometric sequence $\left({t}_{n}\right) = a \cdot {r}^{n - 1}$

So, ${t}_{5} = \left(\sqrt{2} - 1\right) \cdot {\left(\sqrt{2} + 1\right)}^{5 - 1}$

$= \left(\sqrt{2} - 1\right) \left(\sqrt{2} + 1\right) {\left(\sqrt{2} + 1\right)}^{3}$

$= \left[{\left(\sqrt{2}\right)}^{2} - {1}^{2}\right] \left[{\left(\sqrt{2}\right)}^{3} + 3 \cdot \left({\sqrt{2}}^{2}\right) \cdot 1 + 3 \cdot \sqrt{2} \cdot {1}^{2} + {1}^{3}\right]$

$= \left(2 - 1\right) \left(2 \sqrt{2} + 6 + 3 \sqrt{2} + 1\right) = 7 + 5 \sqrt{2}$

May 26, 2018

$x = 2 \mathmr{and} {5}^{t h} \text{ term} = 7 + 5 \sqrt{2}$.

#### Explanation:

For any $3$ consecutive terms $a , b , c$ of a GP, we have,

${b}^{2} = a c$.

Hence, in our case, ${1}^{2} = \left(\sqrt{x} - 1\right) \left(\sqrt{x} + 1\right) = {\left(\sqrt{x}\right)}^{2} - {1}^{2} ,$

$i . e . , 1 = x - 1 , \mathmr{and} , x = 2$.

With $x = 2$, the ${1}^{s t} \mathmr{and} {2}^{n d}$ terms of the GP under

reference are, $\sqrt{x} - 1 = \sqrt{2} - 1 \mathmr{and} 1$, resp.

So, the common ratio r=(2^(nd)" term)"-:(1^(st)" term)",

$= \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$.

:. 4^(th)" term=r("3^(rd)" term)=(sqrt2+1)(sqrtx+1),

$= \left(\sqrt{2} + 1\right) \left(\sqrt{2} + 1\right)$,

$= 2 + 2 \sqrt{2} + 1$,

$= 3 + 2 \sqrt{2}$.

Further, $\left({5}^{t h} \text{ term)=r(} {4}^{t h} t e r m\right)$,

$= \left(\sqrt{2} + 1\right) \left(3 + 2 \sqrt{2}\right)$,

$= 3 \sqrt{2} + 3 + 2 \sqrt{2} \cdot \sqrt{2} + 2 \sqrt{2}$.

$\Rightarrow {5}^{t h} \text{ term} = 7 + 5 \sqrt{2}$.