If 1/q+r, 1/r+p, 1/p+q are in AP, then proof that, p², q², r² are in AP
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If 1/q+r, 1/r+p, 1/p+q are in AP, then proof that, p², q², r² are in AP
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Step-by-step explanation:
1/q+r, 1/r+p, 1/p+q are in AP
then....
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)(r+p)]
(q - p)/[(r+p) (q+r)] = (r-q) [(p+q)(r+p)]
(q-p)/ (q+r) = (r-q)/ (p+q)
(q-p) (p+q) = (r+q)(r-q)
q² - p² = r² - q²
2q² = p² + r²
hence....
p², q² & r² are in AP