Ex. 21. If the sum of m terms of an A. P. is equal to half the sum of
(m+n) terms and is also equal to half the sum of (m +p) terms, prove that
(m+n) (1/m-1/p) =(m+p) (1/m-1/n)
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Answer:
Sm=m/2[2a+(m−1)d]
Sn'=m+n/2[2a+(m+n−1)d]−m/2[2a+(m−1)d]
Sp'=m+P/2[2a+(m+p−1)d]−m/2[2a+(m−1)d]
m/2[2a+(m−1)d]=m+n/2[2a+(m+n−1)d]−m/2[2a+(m−1)d]
2m[2a+(m−1)d]=m+n[2a+(m+n−1)d]
2m/m+n=2a+(m+n−1)d2/a+(m−1)d−1
m−n/m+n=mx+nd−d−md+d/2a+(m−1)d
m−n/m+n=nd/2a+(m−1)d−(1)
m−p/m+p=pd/2a+(m−1)d−(2)
(m−n)(m−p)/(m+n)(m−p)=n/p
m−n/nm(m+p)=(m+n)(m−p)/mp
(m+p)[1/n−1/m]=(m+n)[1/p−1/m]
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