1well i think you have made a mistake.sum of m terms is equal to next n terms not m+nterms
If in an AP,the sum of m terms is equal to sum of next n terms as well as sum of next p terms,then prove:
(m+n){(1/m) -(1/p)}=(m+p){(1/m)-(1/n)}
My attempt:
from given information we can say that
2m{2a+(m-1)d}=(m+n){2a+(m+n-1)d}=(m+p){2a+(m+p-1)d}
after this,i am unable to complete the solution but just have a feeling that we have to eliminate a and d.[2]..but no idea on how to eliminate them
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10 Answers
62one suggestion...
divide by a to reduce the number of variables to 1 (namely d/a)
11i have considered it and only after that i have come up with the equations i wrote!
30\frac{m}{2}\left(2a+(m-1)d \right)=\frac{n}{2}\left(2a+(2m+n-1)d \right)=\frac{p}{2}\left(2a+(2m+p-1)d \right)
\Rightarrow m\left(2a+(m-1)d \right)=n\left(2a+(2m+n-1)d \right)
\Rightarrow 2a(m-n)=d(n-m)\left( \frac{2mn}{n-m}+n+m-1\right)
\Rightarrow 2a=d\left( 1-m-n-\frac{2mn}{n-m}\right) ... (i)
Similarly,
2a=d\left( 1-m-p-\frac{2mp}{p-m}\right) ... (ii)
From (i) and (ii), we get,
m+p+\frac{2mp}{p-m}=m+n+\frac{2mn}{n-m}
\Rightarrow p+\frac{2}{\frac{1}{m}-\frac{1}{p}}=n+\frac{2}{\frac{1}{m}-\frac{1}{n}}
\Rightarrow 2\left(\frac{1}{n}-\frac{1}{p} \right)=\left(\frac{1}{m}-\frac{1}{p} \right)\left(\frac{1}{m}-\frac{1}{n}\right)\left(p-n \right)
\Rightarrow 2\left(\frac{1}{np}\right)=\left(\frac{1}{m}-\frac{1}{p} \right)\left(\frac{1}{m}-\frac{1}{n}\right)
\Rightarrow 2=\frac{(p-m)(n-m)}{m^{2}}
\Rightarrow m=\frac{pn}{m}-p-n
\Rightarrow (m+p)\left(\frac{1}{m}-\frac{1}{n} \right) =2=(m+n)\left(\frac{1}{m}-\frac{1}{p} \right)
Phew! Always knew it would get dirty, hope I made no stupid mistake here. I hope we find a better solution!
1can u explain the first equation u wrote?...i can't get how u got it.
30Sum of next n terms = \frac{n}{2}\left(a_{m+1}+a_{m+n} \right)=\frac{n}{2}\left(a+md+a+(m+n-1)d \right)=\frac{n}{2}\left(2a+(2m+n-1)d \right)
Similarly, for the next p terms.
1*bows*
now i realized y i failed at doing this one...i used sum of next n terms=S(m+n)-S(m)[3][3]...
well can anyone solve it using this way?
1@ashish
isn't the last term of your first step p2 (2a + (2m+2n+p-1)d)
please do check.
30Nope. It is what I've written. The next p terms here mean the next p terms from m.