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The sums of n terms of two A.P.'s are in the ratio 5n + 4 : 9n + 6 . Find the ratio of their 18th terms​​​​

Posted by Umang Kumar (Jun 24, 2017 5:39 a.m.) (Question ID: 6140)

• Let the Sum of nth term of first AP be {tex}{{\rm{S}}_n}{/tex} and that of second term be S'n.

=>     {tex}{{{n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]} \over {{n \over 2}\left[ {2a' + \left( {n - 1} \right)d'} \right]}} = {{5n + 4} \over {9n + 6}}{/tex}

=>     {tex}{{2a + \left( {n - 1} \right)d} \over {2a' + \left( {n - 1} \right)d'}} = {{5n + 4} \over {9n + 6}}{/tex}

Putting n = 35, we get

=>     {tex}{{2a + \left( {35 - 1} \right)d} \over {2a' + \left( {35 - 1} \right)d'}} = {{5 \times 35 + 4} \over {9 \times 35 + 6}}{/tex}

=>     {tex}{{2a + 34d} \over {2a' + 34d'}} = {{175 + 4} \over {315 + 6}}{/tex}

=>     {tex}{{a + 17d} \over {a' + 17d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{a + \left( {18 - 1} \right)d} \over {a' + \left( {18 - 1} \right)d'}} = {{179} \over {321}}{/tex}

=>     {tex}{{{a_{18}}} \over {a{'_{18}}}} = {{179} \over {321}}{/tex}

Therefore, the ratio of 18th term of two APs is 179 : 321.

Answered by Rashmi Bajpayee (Jun 24, 2017 12:44 p.m.)
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If f(x)=(a-xn)1/n,a Is greater than 0 and n belongs to N then prove that f(f(x))=x for all x.

<hr>

Posted by Pfff Xy (Jun 22, 2017 9:27 p.m.) (Question ID: 6108)

If f is a real function defined by f(x)=x-1/x+1,then prove that : f(2x)=3f(x)+1/f(x)+3

Posted by Pfff Xy (Jun 22, 2017 9:11 p.m.) (Question ID: 6106)

Convert 4 Radian into degree measure

Posted by Sarthak Chauhan (Jun 22, 2017 6:33 p.m.) (Question ID: 6101)

• Radian measure = {tex}{\pi\over 180}\times degree\ measure{/tex}

So,

Degree measure = {tex}{18\over \pi}\times radian \ measure{/tex}

=> Degree measure = {tex}{180\over 22}\times 7\times 4{/tex}

= 229.09

Answered by Payal Singh (Jun 22, 2017 7:11 p.m.)
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Prove that cos^2A+cos^2(A+120)+cos^2(A-120)=3/2

Posted by Himanshu Goyal (Jun 16, 2017 11 p.m.) (Question ID: 5961)

• {tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {3 \over 2}{/tex}

We know that

{tex}{\cos ^2}{\rm{A}} = {{1 + \cos 2{\rm{A}}} \over 2}{/tex}                                    ..............(i)

{tex}{\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) = {{1 + \cos \left( {{\rm{2A}} + {{240}^ \circ }} \right)} \over 2}{/tex}         ..............(ii)

And   {tex}{\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {{1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}..............(iii)

Adding all these three equations,

{tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right){/tex} = {tex}{3 \over 2} + {{1 + \cos 2{\rm{A}}} \over 2} + {{1 + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right)} \over 2} + {{1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 1 + \cos 2{\rm{A + }}1 + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right) + 1 + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} + \cos \left( {{\rm{2A + }}{{240}^ \circ }} \right) + \cos \left( {{\rm{2A}} - {{240}^ \circ }} \right)} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} + 2\cos 2{\rm{A}}.\cos {{240}^ \circ }} \over 2}{/tex}

= {tex}{{3 + 3 + \cos 2{\rm{A}} - \cos 2{\rm{A}}} \over 2}{/tex}

= {tex}{6 \over 2}{/tex}

{tex}\Rightarrow{/tex} {tex}{\cos ^2}2{\rm{A}} + {\cos ^2}2\left( {{\rm{A}} + {{120}^ \circ }} \right) + {\cos ^2}2\left( {{\rm{A}} - {{120}^ \circ }} \right) = {3 \over 2}{/tex}

Answered by Rashmi Bajpayee (Jun 17, 2017 noon)
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Q.1 If θ is the positive acute angle , then solve the equation
4 cos2 θ - 4 sinθ = 1 .

Q.2 Find the value of sin 2θ , when tan θ = 247 and θ lies in 3rd quadrant.

Q.3 If sinθ =  1213 , then find the quadrants in which θ can lie. Also , find the values of the remaining trigonometric ratios of θ .

Q.4 A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88m, when it has traced out 720 at the centre. Then, find the length of rope.

Posted by Abhishek Kapoor (Jun 16, 2017 12:26 p.m.) (Question ID: 5935)

• It is 24 /7 in question no. 2

Posted by Abhishek Kapoor (Jun 16, 2017 2:10 p.m.)
• It is 12/13 in question no. 3

Posted by Abhishek Kapoor (Jun 16, 2017 2:09 p.m.)
• Q.2     {tex}\sin 2\theta = {{2\tan \theta } \over {1 + {{\tan }^2}\theta }}{/tex}

Since {tex}\theta {/tex} is in third quadrant, therefore value of tangent is positive.

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{2 \times {{24} \over 7}} \over {1 + {{\left( {{{24} \over 7}} \right)}^2}}}{/tex}

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{{{48} \over 7}} \over {1 + {{576} \over {49}}}}{/tex}

{tex}\Rightarrow{/tex}     {tex}\sin 2\theta = {{48} \over 7} \times {{49} \over {625}} = {{336} \over {625}}{/tex}

Answered by Rashmi Bajpayee (Jun 17, 2017 4:55 p.m.)
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If theta is positive acute angle then solve the equation:--

4cos^2 theta -- 4 sin theta =1

Posted by Abhishek Kapoor (Jun 16, 2017 12:05 p.m.) (Question ID: 5934)

• {tex}4{\cos ^2}\theta - 4\sin \theta = 1{/tex}

=>          {tex}4\left( {1 - {{\sin }^2}\theta } \right) - 4\sin \theta = 1{/tex}

=>          {tex}4 - 4{\sin ^2}\theta - 4\sin \theta = 1{/tex}

=>          {tex} - 4{\sin ^2}\theta - 4\sin \theta + 4 - 1 = 0{/tex}

=>          {tex}4{\sin ^2}\theta + 4\sin \theta - 3 = 0{/tex}

=>          {tex}4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0{/tex}

=>          {tex}2\sin \theta \left( {2\sin \theta + 3} \right) - 1\left( {2\sin \theta + 3} \right) = 0{/tex}

=>          {tex}\left( {2\sin \theta + 3} \right)\left( {2\sin \theta - 1} \right) = 0{/tex}

=>          {tex}2\sin \theta + 3 = 0{/tex}  or   {tex}2\sin \theta - 1 = 0{/tex}

=>          {tex}\sin \theta = {{ - 3} \over 2}{/tex}  or  {tex}\sin \theta = {1 \over 2}{/tex}

But {tex}\theta {/tex} is positive, therefore, taking

{tex}\sin \theta = {1 \over 2}{/tex}

=>        {tex}\sin \theta = \sin {30^ \circ }{/tex}

=>        {tex}\theta = {30^ \circ }{/tex}

Answered by Rashmi Bajpayee (Jun 17, 2017 2:14 p.m.)
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convert -3i into polar form.

Posted by Anupam Bhatt (Jun 10, 2017 5:40 p.m.) (Question ID: 5835)

• Let {tex}-3i= 0-3i=r\left( cos\theta +isin\theta \right) {/tex}

Comparing real and imaginary parts

{tex}rcos\theta =0.......................\left( i \right) \\ rsin\theta =-3....................\left( ii \right) \\ {/tex}

Squaring and adding equations (i) and (ii)

{tex}{ r }^{ 2 }{ cos }^{ 2 }\theta +{ { r }^{ 2 } }sin^{ 2 }\theta =9\\ \Rightarrow { r }^{ 2 }=9\quad \quad \quad \quad \quad \quad \quad \left( { \because cos }^{ 2 }\theta +sin^{ 2 }\theta =1 \right) {/tex}

{tex}\Rightarrow r=3{/tex}

Substituting {tex}r=3{/tex} in equations (i) and (ii), we get

So {tex}\theta {/tex} lies in fourth quadrant

{tex}\therefore \quad \theta =\frac { -\Pi }{ 2 } {/tex}     [ in the fourth quarant format of amplitude or principal argument is {tex}-\theta {/tex} (where {tex}\theta =\frac { \Pi }{ 2 } {/tex} ) ]

{tex}-3i=3\left( cos\left( \frac { -\Pi }{ 2 } \right) +isin\left( \frac { -\Pi }{ 2 } \right) \right) \\ \Rightarrow -3i=3\left( cos\left( \frac { \Pi }{ 2 } \right) -isin\left( \frac { \Pi }{ 2 } \right) \right) \quad \quad \left[ \because cos\left( -\theta \right) =cos\theta \quad ,sin\left( -\theta \right) =-sin\theta \right] {/tex}

Answered by Rasmi RV (Jun 12, 2017 8:50 a.m.)
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now days 11th class mathatics follow modern abc book so ur solution ncert not sufficient so plz update with modern book solution

Posted by gurpreet singh (May 21, 2017 7:30 a.m.) (Question ID: 5402)

If a,b,c € R, then find the relation between a,b&c such that (b-x)²-4(a-x)(c-x)=0 has real & equal roots.

Posted by satvik kumar (May 04, 2017 12:41 p.m.) (Question ID: 5106)

Find domain of1/(x-4)(x-1)

Posted by Raj Yadav (May 03, 2017 5:42 p.m.) (Question ID: 5093)

• Ans. {tex}f(x) = {1\over (x-4)(x-1)}{/tex}

We need to find where the expression is undefined. These values are not part of the domain.

For the above function, if it is defined then

{tex}(x-4)(x-1) \ne 0 {/tex}

=> {tex}x-4\ne 0 \space and\space x-1\ne 0 {/tex}

{tex}x\ne 4 \space and\space x\ne 1 {/tex}

So Domain of f(x) = R - {1,4}

Answered by Naveen Sharma (May 04, 2017 9:41 a.m.)
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Find domain of(x-4)(x-1)?

Posted by Raj Yadav (May 03, 2017 5:41 p.m.) (Question ID: 5092)

• {tex}\left( {x - 4} \right)\left( {x - 1} \right) = {x^2} - 5x + 4{/tex} is a polynomial.

we know that Polynomial is defined for every real x.

hence Domain of {tex}\left( {x - 4} \right)\left( {x - 1} \right){/tex}=R

Answered by Neeraj Sharma (May 06, 2017 9:28 p.m.)
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evaluate {tex}evaluate sigma superscript 13 subscript n=1 i^n + i^n+1) where n belongs to N{/tex}

Posted by Bahirithi Karempudi (Apr 22, 2017 11:23 a.m.) (Question ID: 4991)

Write the following sets in roster form

1) B = {x : x is a positive factor of a prime number}

2) C={x : x2=x where x belongs to Real numbers}

Fast.pls its very urgent..

Posted by Anshika Solanki (Apr 20, 2017 9:11 p.m.) (Question ID: 4975)

Write the following sets in roster form

1) B = {x : x is a positive factor of a prime number}

2) C={x : x2=x where x belongs to Real numbers}

Posted by Anshika Solanki (Apr 19, 2017 10:36 p.m.) (Question ID: 4956)

• Prime numbers have only two factors i.e. 1 and itself

{tex}B = \left\{ {1,x} \right\}{/tex}

As we know

{tex}\eqalign{ & {\left( 0 \right)^2} = 0 \cr & {\left( 1 \right)^2} = 1 \cr} {/tex}

{tex}C = \left\{ {0,1} \right\}{/tex}

Answered by Neeraj Sharma (May 06, 2017 9:37 p.m.)
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If AunionB=AintersectionB then proved A=B

Posted by Rupam Ghosh (Apr 11, 2017 4:12 p.m.) (Question ID: 4833)

F(x) = x÷1+x²

Posted by rishab mittal (Mar 30, 2017 8:58 p.m.) (Question ID: 4620)

• X÷1+x*

2x^

Answered by Abhishek Bhardwaj (Mar 31, 2017 7:41 a.m.)
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if f is a function satisfying f(x+y)=f(x) f(y) for all x,y belongs to n such that f(1)=3 and sigma x =1 to n f(x)=120 then find n.

Posted by Koratamaddi Vamsi (Mar 30, 2017 7:03 p.m.) (Question ID: 4617)

• Ans. Given: f(x + y) = f(x)*f (y) for all x, y {tex}\in {/tex} N ......(1)

{tex}\displaystyle\sum_{i=1}^{n} f(x) = 120 \space \space ............(2){/tex}

f(1) = 3

Taking x = y = 1 in (1), we obtain

f(1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly, f(3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f(4) = f(1 + 3) = f (1) f (3) = 3 × 27 = 81

f(1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P.

First term is 3 and common ratio is also 3.

Also we know

{tex}=> S_n = {a(r^n -1)\over r - 1}{/tex}

{tex}=> S_n = {3(3^n -1)\over 3 - 1}{/tex}  .................... (3)

From (2) and (3), We get

{tex}=> 120 = {3(3^n -1)\over 3 - 1}{/tex}

{tex}=> 120 = {3(3^n -1)\over 2}{/tex}

{tex}=> 80 = 3^n -1{/tex}

{tex}=> 81 = 3^n {/tex}

{tex}=> 3^4 = 3^n {/tex}

n = 4,

So Value of n = 4

Answered by Naveen Sharma (Mar 31, 2017 4:53 p.m.)
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please what is the  meaning of general and pricipal solution in trignometry.

Posted by Racheal Asamoah (Mar 29, 2017 8:26 a.m.) (Question ID: 4580)

4 cards are drawn at a time from a pack of 52 cards find the probability of getting all the 4 cards of the same suit.

Posted by Tanu Shree (Mar 24, 2017 8:17 a.m.) (Question ID: 4474)

How to find the ratio  in which the line segment forming the points (4,8,10) and (6,10, -8) is divided by yz- plane

Posted by Sita Kadian (Mar 14, 2017 7:04 p.m.) (Question ID: 3782)

How to find the square root of -7-24i ?

Posted by Sita Kadian (Mar 14, 2017 6:59 p.m.) (Question ID: 3781)

• Ans.

Let {tex}\sqrt {-7-24i} = x+yi{/tex}

Squaring Both Sides,

=> -7-24i = x2 - y+ 2xyi

on comparing, we get

=> x2 - y= -7   ……(1)

and 2xy = -24 ………(2)

We know

(x2+y2)= (x2-y2)2 + 4x2y2

=> (x2+y2)2 = 49 + 576

=> (x2+y2)= 625

=> x2+ y2 = 25  …………(3)

solving (2) and (3), We get

{tex}x = \pm 3 \space \space \space and \space y =\pm4{/tex}

from (2) as product of xy is -ve, so it means x and y are of opposite sign.

So, when x = 3, y = -4

when x = -3, y = 4

Therefore,

{tex}\sqrt {-7-24i} = \pm (3-4i){/tex}

Answered by Naveen Sharma (Mar 16, 2017 9:42 a.m.)
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Find the term independent of x in the expansion (x/√3)+(√3/2x²)^10

Posted by Gurleen Kaur (Mar 11, 2017 9:04 p.m.) (Question ID: 3544)

• your question is worng when i am solved then i find r=10/3 which is not natural

corrwct question is {tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

Posted by Neeraj Sharma (Mar 12, 2017 1:29 p.m.)
• Please answer this question I need it urgently

Posted by Gurleen Kaur (Mar 12, 2017 1:16 p.m.)
• {tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

{tex}{T_{r + 1}}{ = ^{10}}{C_r}{\left( {\sqrt {{x \over 3}} } \right)^{10 - r}}{\left( {{3 \over {2{x^2}}}} \right)^r}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{5 - {r \over 2} - 2r}}{\left( 3 \right)^{ - 5 + {r \over 2} + r}}{\left( 2 \right)^{ - r}}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{{{10 - 5r} \over 2}}}{\left( 3 \right)^{{{3r - 10} \over 2}}}{\left( 2 \right)^{ - r}}{/tex}

{tex}for\,independent\,of\,x{/tex}

{tex}{\left( x \right)^{{{10 - 5r} \over 2}}} = {x^0}{/tex}

{tex}{{10 - 5r} \over 2} = 0{/tex}

{tex}10 - 5r = 0{/tex}

{tex}r = 2{/tex}

{tex}independent\,term\,{T_3}{ = ^{10}}{C_2}{\left( 3 \right)^{ - 2}}{\left( 2 \right)^{ - 2}}{/tex}

{tex} = {{10!} \over {2!8!}} \times {1 \over {36}}{/tex}

{tex} = {{10 \times 9} \over {2 \times 36}}{/tex}

{tex} = {5 \over 4}{/tex}

Answered by Neeraj Sharma (Mar 12, 2017 1:47 p.m.)
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Find Mode-

20-24

24-29

30-34

35-39

40-44

Posted by Renu Gupta (Mar 07, 2017 10:23 p.m.) (Question ID: 3261)

• What is the frequency?

Posted by Manish Gandhi (Mar 08, 2017 12:48 p.m.)

find the derivative of sin2x

Posted by Prem Sharma (Mar 06, 2017 9:39 p.m.) (Question ID: 3201)

• {tex}let\,y = \sin 2x{/tex}

{tex}differentiating\,with\,respect\,to\,x{/tex}

{tex}{{dy} \over {dx}} = {d \over {dx}}\left( {\sin 2x} \right){/tex}

{tex} = \cos 2x{d \over {dx}}\left( {2x} \right){/tex}

{tex} = 2\cos 2x{/tex}

Answered by Neeraj Sharma (Mar 12, 2017 1:51 p.m.)
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Find derivative of function f(x)=sinx/sinx-cosx with respect to x.

Posted by Rudra Choudhary (Mar 06, 2017 9:03 p.m.) (Question ID: 3200)

• Ans. {tex}f(x) = {sinx\over sinx-cosx}{/tex}

Using quotient Rule, we get

{tex}f'(x) ={ {(sinx-cosx){d(sinx)\over dx}- sinx{d(sinx-cosx)\over dx}}\over (sinx-cosx)^2}{/tex}

{tex}=> { (sinx-cosx)cosx - sinx(cosx-(-sinx)) \over (sinx-cosx)^2}{/tex}

{tex}=> {sinx cosx - cos^2x -sinxcosx -sin^2x\over (sinx-cosx)^2}{/tex}

{tex}=> {- (cos^2x +sin^2x)\over (sinx-cosx)^2}{/tex}

{tex}=> {-1\over (sinx-cosx)^2}{/tex}

Answered by Naveen Sharma (Mar 06, 2017 10:21 p.m.)
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I have a doubt in chapter 9 ie, sequences and series.

Question. insert 3 nos. b/w 1 & 256 so that the resulting sequence is a GP

Posted by JAHANVI MALHOTRA (Mar 05, 2017 6:58 p.m.) (Question ID: 3126)

• anybody to answer this ques.

Posted by JAHANVI MALHOTRA (Mar 05, 2017 7:48 p.m.)
• {tex}let\,three\,numbers\,b/w\,1\,and\,256\,be\,{G_1},\,{G_2},\,and\,{G_3}{/tex}

{tex}1,{G_1},\,{G_2},\,{G_3},256\,are\,in\,GP{/tex}

{tex}a = 1{/tex}

{tex}l = 256{/tex}

{tex}a{r^4} = 256{/tex}

{tex}{r^4} = 256{/tex}

{tex}r = 4{/tex}

{tex}Now\,{G_1} = ar = 1 \times 4 = 4{/tex}

{tex}{G_2} = a{r^2} = 1 \times {\left( 4 \right)^2} = 16{/tex}

{tex}{G_3} = a{r^3} = 1 \times {\left( 4 \right)^3} = 64{/tex}

Answered by Neeraj Sharma (Mar 12, 2017 2:02 p.m.)
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find limits of limx→infinity (root a+x - root 9/x)

Posted by tushar choudhary (Mar 05, 2017 2:36 p.m.) (Question ID: 3103)

Posted by Neeraj Sharma (Mar 12, 2017 2:07 p.m.)

let f:r-r be a function given by f(x)=x2+2.find f-1(27).

Posted by tushar choudhary (Mar 05, 2017 2:31 p.m.) (Question ID: 3102)

• {tex}let\,{f^{ - 1}}\left( {27} \right) = x{/tex}

{tex} \Rightarrow f\left( x \right) = 27{/tex}

{tex} \Rightarrow {x^2} + 2 = 27{/tex}

{tex} \Rightarrow {x^2} = 25{/tex}

{tex} \Rightarrow x = \pm 5{/tex}

{tex}Hence{/tex}

{tex}{f^{ - 1}}\left( {27} \right) = \left\{ { - 5,5} \right\}{/tex}

Answered by Neeraj Sharma (Mar 12, 2017 2:16 p.m.)
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find two no. such that their arithematic mean is 5 and G.M is 9 without using the identity (a+b)2=(a-b)2+4ab

Posted by tushar choudhary (Mar 05, 2017 2:28 p.m.) (Question ID: 3100)

• There are no two numbers such that their arithmatic mean is 5 and G.M. is 9

Because we have know that

AM>GM for between any two numbers a and  b.

here AM=5 and GM=9

9>5 or GM>AM which is invalid

Answered by Neeraj Sharma (Mar 12, 2017 2:36 p.m.)
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