If first and (2n-1)th terms of an AP,GP and H.P are equal and their n th terms are a,b,c respectively then-
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OpenStudy (kanwal32):
options-
A)a+2b B)a+c=b C)\[a \ge b \ge c\] D)b^2=ac
OpenStudy (kanwal32):
@ganeshie8 help @hartnn @wio @ParthKohli @paki
OpenStudy (kanwal32):
@ganeshie8 pls help
OpenStudy (anonymous):
Stop spamming chat.
OpenStudy (kanwal32):
DO U KNOW THE SOLUTION
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OpenStudy (anonymous):
Yeah.
OpenStudy (kanwal32):
@wio pls help
OpenStudy (anonymous):
Do what everyone else does. Post wait, if it still isn't answered, bump it.
OpenStudy (kanwal32):
@Skrilluh I don't want ur advice
OpenStudy (anonymous):
We don't want your spam.
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OpenStudy (kanwal32):
i asked @wio
OpenStudy (kanwal32):
@ganeshie8 help
OpenStudy (kanwal32):
@mathslover help
OpenStudy (kanwal32):
@wio
OpenStudy (kanwal32):
@ganeshie8
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OpenStudy (anonymous):
*Sigh* if i give you the answer will you stop?
OpenStudy (kanwal32):
yeah sure
OpenStudy (anonymous):
B
OpenStudy (kanwal32):
@skillruh answer is wrong
OpenStudy (kanwal32):
@Skrilluh answer is wrong
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OpenStudy (anonymous):
Okay.
OpenStudy (anonymous):
You must not understand the rules of this site...
OpenStudy (kanwal32):
@ParthKohli
OpenStudy (anonymous):
Are you just listing random users?
Parth (parthkohli):
Note that the \(n\)th term of all the three series is the AM, GM and HM of the first term and the \(2n-1\)th term.
Let the first and \(2n-1\)th terms be \(x,y\).
It is given that:\[\dfrac{x+y}{2} = a\]\[\sqrt{xy} = b\]\[\dfrac{2xy}{x+y}=c\]From this, it is clear that \(b^2 =ac.\)