Question

4)For what values of x, the terms
4/3, x, 4/27 are in G.P?
​

Answer 1

Answer:

Step-by-step explanation:

t1 =4/3, t2=x, t3=4/27.

= r=t2/t1 =x÷4/3

Also, t3/t2=4/27÷x

x×x =4/3=4/27

=x^2=16/81

=x=4/9

Hence the the value for x = 4/9.

Answer 2

The required value of x = 4/9

Given :

4/3 , x , 4/27 are in G.P

To find :

The value of x

Solution :

Step 1 of 2 :

Write down three terms of the GP

Here it is given that the 4/3 , x , 4/27 are in G.P

Step 2 of 2 :

Find the value of x

We know that if a , b , c are in GP then b² = ac

So by the given condition

$\displaystyle \sf{ {x}^{2} = \frac{4}{3} \times \frac{4}{27} }$

$\displaystyle \sf{ \implies {x}^{2} = \frac{16}{81} }$

$\displaystyle \sf{ \implies x = \sqrt{ \frac{16}{81} }}$

$\displaystyle \sf{ \implies x = \frac{4}{9} }$

Hence the required value of x = 4/9

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