1Yup.
That is why a subjective paper can't be replaced by an objective one
An AP and a GP have the same first and last terms as well as the same number of terms.
Is the sum of terms greater for the AP or GP?
-
UP 0 DOWN 0 1 6
6 Answers
62Proof by example!
This is the thing I hate about this new pattern... It makes you to stop thinking beyond examples sometimes... (Which I think is suicidal.. Unless you can solve it the right way.. the examples way will often lead you nowhere... )
162Philip.. but I am not so much against the objective paper either..
I think that for 0those who have studied well the format of paper doesnt matter at all..
1ok.. but still i think only a subjective paper can really test us.
Besides, there is nothing an objective paper can test which a subjective one cannot (except all that beymani and tukkebaji :D )
and my maths teacher is hoping that by 2011 we'll again have a subjective paper. (a bleak prospect)
11For the A.P, let the mth term from the beginning be p, and the mth term from the end be q....
Needless to mention, p+q=a+b
Now for the G.P, we have the sum of equivalent p's and q's as \frac{ar^m}{r}+\frac{b}{r^m}=ar^{m-1}+br^{-m}
Now we need to see which is greater (a+b)....or....(ar^{m-1}+br^{-m} )
Now a dumb & dead proof of the inequality.....f(r)=r-r^m+r^n-r^{n-m}
Derivative gives f'(r)=1-mr^{m-1}+(n-m)r^{n-1}-(n-m)r^{n-m-1}+nr^{n-1}>0 for all r>1.....which can be always taken, as if r<1, then from the reversed end r turns out to be >1......
Thus f(r)>0 for all r>1, and hence Sum of A.P>Sum of G.P
For negative terms, we can similarly proceed by taking the modulo of the terms!
1 of my friends told me that the inequality of f(r)>0 can also be done by Jenson's - I don't know Jenson's inequality, can someone give that proof?
