IIT JEE , AIEEE Forum - Mathematics , Algebra

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Consider the following inequation........

$\sum_{1}^{120}[ r / ( x - r ) ] >= 3$ ....( r varies from 1 to 120 )

The values of x which satisfy the above eqn . are

$x\in$ ( a, b ) union ( c, d ) union ( e , f ) ........................( consider all

the sets of x for which the above holds true )

FIND :: ( b + d + f + ...........) - ( a + c + e + ..... )

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first soln getz 5 salutes...

sriram.a

edited 115

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nopes...

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no replies still....

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shud i reveal d ans..

aditi_g

is it 2420??

akku

is it 60??

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bang...the answer is 2420...proof plz. and u get ur salutes..!!

aditi_g

c for the diff values of r

r=1

a,b = 1,4/3

for r=2

c,d=2,8/3

for r=3

e,f= 3,12/3

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similarly the last pair wud be 120,480/3

so b+d+f....=4/3[1+2+3+...120]

a+c+e....=1+2+3+...120

so b+d+f.... - (a+c+e...)=

4/3(120)(121)/2 - 120(121)/2 [using sum of n terms]

=1/3(120)(121)/2

=2420

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brilliant...u get ur salutes..!!

aditi_g

thanx but i dnt think it was tht tuff was it??

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