Question

find common ratio for GP 1/√2,-2,8/√3​

Answer 1

GiveN :

• 1/√2 , -2 , 8/√3 is a GP

To FinD :

• Common Ratio (r)

SolutioN :

Use formula for common ratio :

$\longrightarrow \sf{r_1 \: = \: \dfrac{a_2}{a_1}} \\ \\ \longrightarrow \sf{r_1 \: = \: \dfrac{\dfrac{-2}{1}}{\sqrt{2}}} \\ \\ \longrightarrow \sf{r_1 \: = \: \dfrac{2 \sqrt{2}}{1}}$

• So, r1 is 2√2/1

______________________________

Now,

$\longrightarrow \sf{r_2 \: = \: \dfrac{a_3}{a_2}} \\ \\ \longrightarrow \sf{r_2 \: = \: \dfrac{\dfrac{8}{\sqrt{3}}}{-2}} \\ \\ \longrightarrow \sf{r_2 \: = \: \dfrac{-2 \: \times \: 8}{\sqrt{3}}} \\ \\ \longrightarrow \sf{r_2 \: = \: \dfrac{-16}{\sqrt{3}}}$

• So, r2 is -16/√3

______________________________

As,

$\longrightarrow \sf{r_1 \: \neq \: r_2}$

So, therefore no common ratio exists

• and the given terms are not in GP

Answer 2

${\bf{\boxed{\bf{\red{Answer:}}}}}$

Given question:

⇏ $\tt \dfrac{1}{ \sqrt{2} } \: ,-2 \: , \dfrac{8}{43} = GP$

To Find:

⇏ $\tt Find \: common \: ratio.$

Note:

⇏ $\tt The \: given \: terms \: aren't \: in \: GP.$

Using formula:

⇏ $\sf r_1 = \dfrac{a_{2}}{a_{1} }$

⇏ $\sf r_{1} = \dfrac{ \dfrac{ - 2}{1} }{ \sqrt{2} }$

⇏ ${\bf{\boxed{\bf r_{1} = \dfrac{2 \sqrt{2} }{1} }}}}$

Therefore, r₁ = 2 √2/1.

⇏ $\sf r_{2} = \dfrac{a_{3}}{a_{2} }$

⇏ $\sf r_{2} = \dfrac{ \dfrac{8}{ \sqrt{3} } }{ - 2}$

⇏ $\sf r_{2} = \dfrac{ - 2 \times 8}{ \sqrt{3} }$

⇏ ${\bf{\boxed{\bf r_{2} = \dfrac{ - 16}{ \sqrt{3} }}}}$

Therefore, r₂ = -16/√3.

Finally, r₁ ≠ r₂ and there's no common ratios.