If the sum of `m` terms of an A.P. is equal to the sum of either the next `n` terms or the next `p` terms, then prove that `(m+n)(1/m-1/p)=(m+p)(1/m-1/n)dot`

Explanation

Let the AP be `a,a+d,a+2d,"…"`
Given,`T_(1)+T_(2)+T_(3)+"......."+T_(m)=T_(m+1)+T_(m+2)+"......"T_(m+n)" " "....(i)"`
On adding `T_(1)+T_(2)+"......."+T_(m)` both sides in Eq. (i), we get
`2(T_(1)+T_(2)+"......."+T_(m))=T_(1)+T_(2)+"......."+T_(m)+T_(m+1)+"......"T_(m+n)`
`implies 2S_(m)=S_(m+n)`
`2*(m)/(2)[2a+(m-1)d]=(m+n)/(2)[2a+(m+n-1)d]`
Let `2a+(m-1)d=x`
`implies mx=(m+n)/(2){x+nd}`
`implies (m-n)x=(m+n)nd " " "......(ii)"` Again, `T_(1)+T_(2)+"......."+T_(m)=T_(m+1)+T_(m+2)+"......."+T_(m+p)` ,brgt Similarly, `(m-p)x=(m+p)pd" ' "....(iii)"`
On dividing Eq. (ii) by Eq. (iii), we get
`(m-n)/(m-p)=((m+n)n)/((m+p)p)`
`implies (m-n)(m+p)p=(m-p)(m+n)n`
On dividing both sides by mnp, we get
`(m+p)((1)/(n)-(1)/(m))=(m+n)((1)/(p)-(1)/(m))`
Hence, `(m+n)((1)/(m)-(1)/(p))=(m+p)((1)/(m)-(1)/(n))`