A number is the reciprocal of the other. If t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two numbers be $a$ and $b$.Given that one number is the reciprocal of the other,so $a = \frac{1}{b}$ or $ab = 1$.The arithmetic mean of th

Vedclass · Accepted Answer

$\frac{3}{2}, \frac{2}{3}$

Vedclass · Answer

(D) Let the two numbers be $a$ and $b$.Given that one number is the reciprocal of the other,so $a = \frac{1}{b}$ or $ab = 1$.The arithmetic mean of the two numbers is given as $\frac{a + b}{2} = \frac{13}{12}$.This implies $a + b = \frac{13}{6}$.Substituting $b = \frac{1}{a}$ into the equation,we get $a + \frac{1}{a} = \frac{13}{6}$.Multiplying by $6a$,we get $6a^2 - 13a + 6 = 0$.Factoring the quadratic equation: $6a^2 - 9a - 4a + 6 = 0 \Rightarrow 3a(2a - 3) - 2(2a - 3) = 0$.$(3a - 2)(2a - 3) = 0$.Thus,$a = \frac{2}{3}$ or $a = \frac{3}{2}$.If $a = \frac{3}{2}$,then $b = \frac{2}{3}$. If $a = \frac{2}{3}$,then $b = \frac{3}{2}$.Therefore,the numbers are $\frac{3}{2}$ and $\frac{2}{3}$.

$A$ number is the reciprocal of the other. If the arithmetic mean of the two numbers is $\frac{13}{12}$,then the numbers are

Similar Questions

If $3 + \frac{1}{4} (3 + d) + \frac{1}{4^2} (3 + 2d) + \dots \infty = 8$,then the value of $d$ is:

The $n^{th}$ term of the series $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ will be

If $f(x + y, x - y) = xy$,then the arithmetic mean of $f(x, y)$ and $f(y, x)$ is

$\sum\limits_{r = 0}^{100} {({r^2} + 4r + 4)(r + 1)!}$ is equal to :-

If $|\alpha| < 1$ and $|\beta| < 1$,where $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty$ and $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots \infty$,then $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module