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luke97 April 23rd, 2014 07:02 AM

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

v8archie April 23rd, 2014 08:09 AM

Quote:

Originally Posted by luke97 (Post 191560)
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.

Denis April 23rd, 2014 05:04 PM

(p - 1)2p = 2p^2 - 2p


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