The value of 3+14+13+14+13+...∞ is:

x=3+14+13+14+13+....∞Then x=3+14+1x=3+x4x+1=13x+34x+1⇒4x2+x=13x+3⇒4x2−12x−3=0⇒x=12±144+488=12±1928=3±232 .x>0⇒x=3+232=1.5+3 .

The value of 3+14+13+14+13+...∞ is:

x=3+14+13+14+13+....∞Then x=3+14+1x=3+x4x+1=13x+34x+1⇒4x2+x=13x+3⇒4x2−12x−3=0⇒x=12±144+488=12±1928=3±232 .x>0⇒x=3+232=1.5+3 .

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The value of $3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + \frac{1}{3 + ... \infty}}}}$ is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$4 + \sqrt{3}$

$3 + 2 \sqrt{3}$

$1.5 + \sqrt{3}$

$2 + \sqrt{3}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$x = 3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + \frac{1}{3 + .... \infty}}}}$

Then $x = 3 + \frac{1}{4 + \frac{1}{x}} = 3 + \frac{x}{4 x + 1} = \frac{13 x + 3}{4 x + 1} \Rightarrow 4 x^{2} + x = 13 x + 3 \Rightarrow 4 x^{2} - 12 x - 3 = 0$ $\Rightarrow x = \frac{12 \pm \sqrt{144 + 48}}{8} = \frac{12 \pm \sqrt{192}}{8} = \frac{3 \pm 2 \sqrt{3}}{2}$ .
$x > 0 \Rightarrow x = \frac{3 + 2 \sqrt{3}}{2} = 1.5 + \sqrt{3}$ .