Question

If 1/q+r, 1/r+p, 1/p+q are in AP, then proof that, p², q², r² are in AP​

Answer 1

Answer:

I hope that it will be helpful for you

Answer 2

Answer:

see below

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