Question

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) ​

Answer 1

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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