Home » Ask & Discuss » Mathematics
« Back to Discussion

Algebra

Cool goIITian

Joined:	8 Nov 2006
Post:	51

14 Nov 2006 21:25:16 IST

0 People liked this

1399

sequences

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.

Comments (7)

kirtana

Cool goIITian

Joined: 8 Nov 2006

Posts: 51

14 Nov 2006 22:09:20 IST

Like 0 people liked this

oh godd
i messed up the forum index
sorry!!!!!!!!

Administrator'

Administrator

Joined: 19 Oct 2006

Posts: 1558

14 Nov 2006 22:26:54 IST

Like 0 people liked this

moving to the correct forum.

~forum administrator

edison

Forum Expert

Joined: 19 Oct 2006

Posts: 7802

15 Nov 2006 16:27:59 IST

Like 0 people liked this

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 (r^n-1 - 1) ...(3)

Substituting the value of (n-1)d from equation (3) in (1), we obtain

X = na + na(r^n-1 - 1)/2 = na[(r^n-1 +1)/2]

So, X = na[(r^n-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.

kirtana

Cool goIITian

Joined: 8 Nov 2006

Posts: 51

15 Nov 2006 23:17:19 IST

Like 0 people liked this

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..

edison

Forum Expert

Joined: 19 Oct 2006

Posts: 7802

19 Nov 2006 16:15:18 IST

Like 0 people liked this

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

hemang

Blazing goIITian

Joined: 27 Dec 2010

Posts: 1508

27 Jan 2011 11:34:55 IST

Like 0 people liked this

good

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2205

27 Jan 2011 18:18:15 IST

Like 0 people liked this

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

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team