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Algebra

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30 Dec 2009 23:10:24 IST

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if am of 5 real numbers a,b,c,d,p is 2 and am of their squares is 4, find the interval in which p li

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if am of 5 real numbers a,b,c,d,p is 2 and am of their squares is 4, find the interval in which p lies.

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Avnish Gaur

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30 Dec 2009 23:28:50 IST

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take 5 no.s as a-2d, a-d, a, a+d, a+2d

so, am=5a/5=2(given)

therefore, a=2,

similiarly, to find am of their squares, square these no.s,

am will be, a(square)+2d(square).

a=2

hence, 4+2d(square)=4

therefore, d=0

hence all numbers are 2

akki ~~ unlucky forever ~~

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30 Dec 2009 23:54:24 IST

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@ avnish . . . it is not given that the 5 no's are in AP . . . .

@ jyotibdha, it must be given in the question that the 5 no's, or atleast the 4 no's 'a,b,c,d' are "positive real" . . . i am taking a,b,c,d, to be positive reals,else the que is unsolvable

$\frac{a+b+c+d+p}{5}=2\\ \\ \\ \frac{a^2+b^2+c^2+d^2+p^2}{5}=4 \\ \\ \\ => a+b+c+d=10-p\ ,\\ \\ a^2+b^2+c^2+d^2=20-p^2\\ \\ \\ now, using\ QM-AM\ ineq.\ to\ a,b,c,d \\ \\ \\ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \frac{a+b+c+d}{4}\\ \\ \\ \sqrt{\frac{20-p^2}{4}} \ge \frac{10-p}{4}\\ \\ \\ solving,(p-2)^2 \le 0 \\ \\ =>(p-2)^2=0\\ \\ \\ => p=2$

jyotibdha acharya

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30 Dec 2009 23:58:48 IST

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i also got that rslt p=2.but the options are

1.[-16/5,16/5]

2.[0,2]

3.[-2,2]

4.[-16/5,0]

akki ~~ unlucky forever ~~

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31 Dec 2009 00:02:52 IST

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yaa p = 2, so p(=2) lies in interval

1.[-16/5,16/5]

2.[0,2]

3.[-2,2]

so 1,2,3 options will be correct . . .

edison

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2 May 2010 17:22:35 IST

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well done Avinash and akki

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