Question

the next term in the ap of 1/1+√x , 1/1-x ,1/1-√x is​

Answer 1

Step-by-step explanation:

The A.P is :$\frac{1}{1 + \sqrt{x} } \frac{1}{1 - x} \frac{1}{1 - \sqrt{x} } ...$

Here :

$a = \frac{1}{1 + \sqrt{x} } \\ d = \frac{1}{1 - x} - \frac{1}{1 + \sqrt{x} } = \frac{1 + \sqrt{x} - 1 + x}{1 + \sqrt{x} - x - x \sqrt{x} } \\ = \frac{x + \sqrt{x} }{ \sqrt{x} - x \sqrt{x} - x } \\ = \frac{ \sqrt{x} ( \sqrt{x} + 1)}{ \sqrt{x}(1 - x - \sqrt{x}) } \\ = \frac{1 + \sqrt{x} }{1 - \sqrt{x} - x }$

Also :

$d = \frac{1 }{1 - \sqrt{x} } - \frac{1}{1 - x} \\ = \frac{ \sqrt{x} }{1 - x}$

As we have got a and d,

$a _{4} = a + 3d \\ = \frac{1}{1 + \sqrt{x} } + 3 \times ( \frac{ \sqrt{3} }{1 - x} ) \\ = \frac{1}{1 + \sqrt{x} } + \frac{3 \sqrt{3} }{1 - x} \\ = \frac{1 - x + 3 \sqrt{3} + 3 \sqrt{3x} }{1 + \sqrt{x} - x - x \sqrt{x} } \\ = \frac{ \sqrt{x } - 3 \sqrt{3} - 1 }{x - 1}$

And here we got the Next term i.e 4th term.