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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+(n-1)d} = n²p => ½{2a+(n-1)d} = np — (i) Given: (m/2){2a+(m-1)d} = m²p => ½{2a+(m-1)d} = mp — (ii) Subtracting equation (i) from (ii), ½{2a+(n-1)d - 2a-d(m-1)} = p(n-m) => ½d(n-m) = (n-m) => d=2p —(iii) Substituiting d by 2p in equation (ii) ½{2a+(m-1)2p} = mp => a=p — (iv) sum of p terms = (p/2) {2a+(p-1)d} = (p/2){2p+(p-1)2p} = p² + p² -p = p(2p-1) 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
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 Originally Posted by luke97 sum of p terms = (p/2) {2a+(p-1)d} = (p/2){2p+(p-1)2p} = p² + p² -p
You have expanded the expression incorrectly. The last line does not follow from what goes before. April 23rd, 2014, 05:04 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 (p - 1)2p = 2p^2 - 2p Tags arithmetic, arithmetic progression, progression Search tags for this page

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