If 1/q+r,1/r+p and 1/p+q are in A.P then show that p^2,q^2 and r^2 are in A.P
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1/(q+r) , 1/(r+p), 1/(p+q) are in AP
it mean common difference always same
so,
1/(r+p) -1/(q+r)=1/(p+q) - 1/(r+p)
(q+r-r-p)/(r+p)(q+r)=(r+p-p-q)/(p+q)(p+r)
(q-p)/(q+r)=(r-q)/(p+q)
use cross multiplication,
q^2-p^2=r^2-q^2
here you see p^2 , q^2 ,r^2 their common difference are same
so, p^2,q^2 ,r^2 are in AP
it mean common difference always same
so,
1/(r+p) -1/(q+r)=1/(p+q) - 1/(r+p)
(q+r-r-p)/(r+p)(q+r)=(r+p-p-q)/(p+q)(p+r)
(q-p)/(q+r)=(r-q)/(p+q)
use cross multiplication,
q^2-p^2=r^2-q^2
here you see p^2 , q^2 ,r^2 their common difference are same
so, p^2,q^2 ,r^2 are in AP
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