If 1/a, 1/b, 1/c are in AP, prove that (i) b + c / a, c +a / b, a + b / c are in AP (ii) a(b +c), b (c +a), c ( a + b) are in AP

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Solution

given: 1/a , 1/b and 1/c are in AP
multiplying each term by (a+b+c) will also result as an AP.
(a+b+c) / a , (a+b+c)/b and (a+b+c)/c must form an AP
subtracting 1 from each term is also an AP
therefore
(a+b+c) / a -1 , (a+b+c)/b -1 and (a+b+c)/c -1 is also an AP.
therefore (b+c)/a , (a+c)/b and (a+b)/c is an AP.
given :
1/a , 1/b and 1/c are in AP
multiplying each term by -abc is also an AP
therefore
-bc , -ca, -ab is also an AP
adding ab+bc+ca to each term, will also an AP
therefore
bc+ca , ab+bc, bc+ca is also an AP
i.e.
c(b+a) , b(a+c) and c(a+b) is also an AP
reverse of the sequence is also an AP
therefore
c(a+b) , b(a+c) and c(a+b) is also an AP

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