sign up I login
 advanced
» win an I-Phone. check i-points

Ask & Discuss Questions with Community & Experts

Moderation Team
Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: progression 1
Forum Index -> Algebra -> View Full Question like the article? email it to a friend.  
Author Message
metal (498)

Blazing goIITian

Olaaa!! Perrrfect answer. 80  [129 rates]

metal's Avatar

total posts: 337    
offline Offline
Very easy!!!!!!!!
 
Take the first term to be a and the last term to be b. Let the common difference be  d .
then, mulitiplying 1st and last you get: ab
and multiplying 2nd and 2nd last you get ab+d(b-a-d).
Now, if d>0, b>a and also b> a+d
so, b-a-d >0
and hence d(b-a-d) > 0
If d<0
b<a
and b<a+d,
so, b-a-d<0
so d(b-a-d)>0
so, product of 2nd and 2nd last = ab+some positive number and is hence greater than ab[product of 1st and last]
So,
ab < ab+d(b-a-d)
so, ab+d(b-a-d)< ab + 2d(b-a-d)
i.e. product of 2nd and 2nd last< product of 3rd and 3rd last.
IN this way the product of equidistant terms keep on incresing as one moves closer to the middle term.
 
please rate me..........
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
 

 Aakash Institute IIT/ AIEEE/ Medical Crash Course
Name  
E-mail  
Phone  
Mobile  
** Hurry. Exclusive goIIT Offer. Limited Seats Only!
available in: New Delhi, Amritsar, Bhatinda, Bokaro, Chandigarj, Dehradun, Guwhati, Hyderabad, Indore, Jaipur, Kanpur, Karnal, Kolkata, Kota, Lucknow, Ludhiana, Mumbai, Noida, Patiala, Patna, Pune, Ranchi, Varanasi
Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Aakash-IITJEE : AIEEE
Aakash-IITJEE : DCE
Aakash-IITJEE : MHTCET
Aakash Institute : AIPMT
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya