progressions-4

i have considered it and only after that i have come up with the equations i wrote!

\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!

can u explain the first equation u wrote?...i can't get how u got it.

Sum 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.

*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?

Nope. It is what I've written. The next p terms here mean the next p terms from m.