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Algebra

Cool goIITian

 Joined: 8 Nov 2006 Post: 51
14 Nov 2006 21:25:16 IST
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Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

hii
i tried a lot for this question but am not reaching newhere.......
An AP and GP with positive terms have the same number of terms and their first and their last terms are equal. show that the sum of the AP is greater than or equal to that of the GP.

Cool goIITian

Joined: 8 Nov 2006
Posts: 51
14 Nov 2006 22:09:20 IST
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oh godd
i messed up the forum index
sorry!!!!!!!!

Joined: 19 Oct 2006
Posts: 1558
14 Nov 2006 22:26:54 IST
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moving to the correct forum.

Forum Expert
Joined: 19 Oct 2006
Posts: 7804
15 Nov 2006 16:27:59 IST
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Let us the consider the following 'n' terms of AP,
a, a+d, a+2d, a+3d,...., [a+(n-1)d] = b
Similarly GP can be considered as
a, ar, ar2, ar3,..., [ar(n-1)] = b
thus, here First and Last terms of AP and GP are same being 'a' and 'b' respectively
now sum of AP is given by
X = na + dn(n-1)/2                               ...(1)
Sum of GP can be written as
Y = a[(rn -1)/(r - 1)]                            ...(2)
Last term of AP = Last term of GP
So, na + dn(n-1)/2 =  a[(rn -1)/(r - 1)]
Simplification of which gives,
d(n-1) = a (rn-1 - 1)                                ...(3)
Substituting the value of (n-1)d from equation (3) in (1), we obtain
X = na + na(rn-1 - 1)/2   = na[(rn-1 +1)/2]

So, X = na[(rn-1 +1)/2]   and    Y = a[(rn -1)/(r - 1)]
Here all terms of AP and GP are positive which means that common ratio of GP  r > 0
Now by the principle of mathematical induction u can prove that X is either equal to or greater than Y.

Cool goIITian

Joined: 8 Nov 2006
Posts: 51
15 Nov 2006 23:17:19 IST
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sir,
we kno that last term of the AP = last term of the GP
then how did u get that
sum of AP = sum of GP..
really stumped..

Forum Expert
Joined: 19 Oct 2006
Posts: 7804
19 Nov 2006 16:15:18 IST
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you can also try approach of mathematical induction.
you have mentioned that the first and last terms are equal therefore if u consider
a, a, a then the three terms are in A.P with common difference 0 as well as in G.P with a common ratio of 1. in this case the sum of AP comes equal to that of GP

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
27 Jan 2011 11:34:55 IST
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good

Forum Expert
Joined: 28 Feb 2007
Posts: 2217
27 Jan 2011 18:18:15 IST
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Let the number of terms be n. Then the sum of terms of the AP is

Suppose the common ratio of the terms is r, then

Now the sum of terms of the GP can be written as

The typical bracketed term is

We will now prove that this is less than or equal to a+b =

WLOG we may assume that r>1 (otherwise we read the sequence backwards)

So we are to prove that

Hence, each bracketed term is less than or equal to a+b.

Hence the entire sum is less than or equal to  thus proving the statement

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