The nth term of the sequence 1-1-x- 1-1-x- 1-1-x-is

Find the termGiven series 1/1+√(x),1/1 x,1/1 √(x),.......Here, the difference d=1/1 x 1/1+√(x)=1/1 √(x) 1/1 x0ex0ex =1 1+√(x)/1 x=1+√(x) 1/1 x0ex0ex =√(x)/1 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

The nth term ...

Question

The $n^{t h}$ term of the sequence $\frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}, . . . . . . .$ is

$\frac{1 + \sqrt{x} (n^{2} - 2)}{1 - x}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1 + \sqrt{x} (n - 1)}{1 + \sqrt{x}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1 + \sqrt{x} (n - 2)}{1 - x}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{3 - \sqrt{x} (n + 2)}{3 (1 - x)}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$\frac{1 + \sqrt{x} (n - 2)}{1 - x}$

Find the term
Given series $\frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}, . . . . . . .$
Here, the difference
$d = \frac{1}{1 - x} - \frac{1}{1 + \sqrt{x}} = \frac{1}{1 - \sqrt{x}} - \frac{1}{1 - x} = \frac{1 - 1 + \sqrt{x}}{1 - x} = \frac{1 + \sqrt{x} - 1}{1 - x} = \frac{\sqrt{x}}{1 - x} = \frac{\sqrt{x}}{1 - x} T_{n} = a + (n - 1) d$
Here, $a = \frac{1}{1 + \sqrt{x}}$
$T_{n} = \frac{1}{1 + \sqrt{x}} + (n - 1) \frac{\sqrt{x}}{1 - x} = \frac{(n - 2) \sqrt{x} + 1}{1 - x}$
Hence, the correct option is $Option (C)$

Suggest Corrections

Similar questions

Q. Consider the sequence

a_{n}

given by

a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}

Let

S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . . . . . . . . . + \frac{1}{a_{n} + 1}

then find the value of

[S_{2012}]

, where [.] denotes greatest integer function.

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. If

ω

is complex cube root of unity and a, b, c are three real numbers such that

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

and

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω

then

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

Q. If

\frac{x}{a} + \frac{y}{b} = 1

is a variable line where

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

(

c

-constant) , then locus of the foot of the perpendicular drawn from origin is

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

Join BYJU'S Learning Program

The nth term of the sequence 11+x,11-x,11-x,....... is

The correct option is C 1+xn-21-xFind the termGiven series 11+x,11-x,11-x,.......Here, the difference d=11-x-11+x=11-x-11-x=1-1+x1-x=1+x-11-x=x1-x=x1-xTn=a+n-1dHere, a=11+xTn=11+x+n-1x1-x=n-2x+11-xHence, the correct option is OptionC

The $n^{t h}$ term of the sequence $\frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}, . . . . . . .$ is