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April 23rd, 2014, 07:02 AM  #1 
Newbie Joined: Apr 2014 From: Rourkela, India Posts: 2 Thanks: 0  Arithmetic Progression
Question— If sum of n terms = n²p and sum of m terms = m²p, in an Arithmetic progression, m≠n, prove that sum of p terms = p³. Well I tried to solve this way. Given: (n/2){2a+(n1)d} = n²p => ½{2a+(n1)d} = np — (i) Given: (m/2){2a+(m1)d} = m²p => ½{2a+(m1)d} = mp — (ii) Subtracting equation (i) from (ii), ½{2a+(n1)d  2ad(m1)} = p(nm) => ½d(nm) = (nm) => d=2p —(iii) Substituiting d by 2p in equation (ii) ½{2a+(m1)2p} = mp => a=p — (iv) sum of p terms = (p/2) {2a+(p1)d} = (p/2){2p+(p1)2p} = p² + p² p = p(2p1) I'm unable to solve further. Please help Last edited by luke97; April 23rd, 2014 at 07:07 AM. 
April 23rd, 2014, 08:09 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra  
April 23rd, 2014, 05:04 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,344 Thanks: 1024 
(p  1)2p = 2p^2  2p


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