Maths-

General

Easy

Question

If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

1:3
1:6
1:9
1:12

hint Hint:

We know that A.M is Arithmetic Mean and G.M is Geometric Mean and the A.M and G.M for two positive numbers, say a and b will be $fraction numerator a plus b over denominator 2 end fraction$ $\frac{a + b}{2}$ and $\sqrt{a b}$ respectively. We will make a quadratic equation using it and the roots of the equation gives the value of the numbers which we have to prove.

The correct answer is: 1:9

Detailed Solution $square root of a b end root$

AM between two positive numbers is given by $fraction numerator a plus b over denominator 2 end fraction$
GM between two positive numbers a and b is given by $square root of a b end root$
We have been given that :
$A plus G equals a minus b S u b s t i t u t i n g space t h e space v a l u e s rightwards double arrow fraction numerator a plus b over denominator 2 end fraction space plus space square root of a b end root space equals space a minus b T a k i n g space L C M rightwards double arrow fraction numerator a space plus space b space plus space 2 square root of a b end root over denominator 2 end fraction space equals a space minus space b rightwards double arrow a space plus space b space plus space 2 square root of a b end root space equals space 2 a space minus space 2 b rightwards double arrow 2 square root of a b end root space equals space a space plus 3 b S q u a r i n g space o n space b o t h space s i d e s rightwards double arrow 4 a b equals space a squared space plus space 9 b squared space minus space 6 a b rightwards double arrow 0 equals space a squared space plus 9 space b squared space minus space space 10 a b D i v i d i n g space b y space a squared space comma space w e space g e t rightwards double arrow 1 space plus 9 space b squared over a squared space minus space space 10 b over a space equals space 0 L e t space b over a space equals space t comma space t h e space e q u a t i o n space b e c o m e s rightwards double arrow 9 space t squared minus space space 10 t space space plus 1 space equals space 0 U sin g space q u a d r a t i c space f o r m u l a space f i n d space r o o t s rightwards double arrow t equals space fraction numerator negative b space plus-or-minus square root of b squared space minus 4 a c end root over denominator 2 a end fraction space equals space fraction numerator 10 space plus-or-minus square root of 100 space minus 36 end root over denominator 18 end fraction space space equals space fraction numerator 10 space plus-or-minus 8 over denominator 18 end fraction rightwards double arrow t equals space 1 comma space 1 over 9 W e space k n o w space t h a t space t space equals space b over a rightwards double arrow b over a equals space 1 space o r space 1 over 9 rightwards double arrow b over a space c a n n o t space b e space 1 space left parenthesis S i n c e space A M space plus space G M space i s space n o t space 0 right parenthesis T h u s comma space b over a space equals space space 1 over 9$

We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving. You can also remember the statement to be proved, given in the question as a property for two positive numbers.

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If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

1:3 1:6 1:9 1:12

We know that A.M is Arithmetic Mean and G.M is Geometric Mean and the A.M and G.M for two positive numbers, say a and b will be and √ respectively. We will make a quadratic equation using it and the roots of the equation gives the value of the numbers which we have to prove.

The correct answer is: 1:9

Detailed SolutionAM between two positive numbers is given by GM between two positive numbers a and b is given by We have been given that :

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1:3
1:6
1:9
1:12

$left parenthesis 666 horizontal ellipsis text n digits end text right parenthesis squared plus left parenthesis 888 horizontal ellipsis n$ digits )=

$left parenthesis 666 horizontal ellipsis text n digits end text right parenthesis squared plus left parenthesis 888 horizontal ellipsis n$ digits )=

In a $increment blank P Q R$ as shown in figure given that $x colon y colon z equals 2 colon 3 colon 6$ , then the value of $angle Q P R$ is

In a $increment blank P Q R$ as shown in figure given that $x colon y colon z equals 2 colon 3 colon 6$ , then the value of $angle Q P R$ is

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?