Question

3. Find the next term of sequence
1/1+√x , 1/1-x , 1/1-√x.......​

Answer 1

Answer:

t(2)—t(1)

=1/(1—x)—1/(1+√x)

=1/(1—x)—[(1—√x)/(1—x)]

=√x

t(3)—t(2)

=1/(1—√x)—1/(1—x)

=[(1+√x)/(1—x)]—1/(1—x)

=√x

this series is an AP

for next term :-

n=4

d=$\sqrt{x}$

t(n)=a+(n-1)d

t(4)=1/($\sqrt{x}$+1)+(4-1)$\sqrt{x}$

Step-by-step explanation:

This is an AP series. simply use the formula of t(n).

Answer 2

The next term of the sequence is equal to (1+2√x)/(1-x).

t(1) - first term of the sequence = 1/(1+√x)

t(2) - first term of the sequence = 1/(1-x)

t(3) - first term of the sequence = 1/(1-√x)

t(2) - t(1) = 1/(1-x) - 1(1+√x)

=> 1/(1-x) - (1-√x)/(1- x)

=> √x/(1-x)

t(3) - t(2) = 1/(1-√x) - 1/(1-x)

=> (1+√x)/(1-x) - 1/(1-x)

=> √x/(1-x)

Since the difference between any two consecutive terms of the sequence is equal , so the sequence is an A.P .

The common difference of A.P = d = √x/(1-x)

first term = a = 1/(1+√x)

The next term(4th term) of the A.P = a + 3d

= 1/(1+√x) + 3√x/(1-x)

=> (1-√x)/(1-x) + 3√x/(1-x)

=> (1+2√x)/(1-x)