# If a, b and c are in H.P, then a/(b+c), b/(c+a), c/(a+b) are in

$(a)\;AP\qquad(b)\;HP\qquad(c)\;GP\qquad(d)\;None\;of\;the\;above$

Explanation : Given that a,b,c are in HP
Therefore 1/a,1/b,1/c are in AP
Multiplying by (a+b+c),
(a+b+c)/a(a+b+c)/b,(a+b+c)/c are in AP
Reducing further
1+(b+c)/a,1+(a+c)/b,1+(a+b)/c are in AP
Minus 1 from each term,
(b+c)/a,(a+c)/b,(a+b)/c are in AP
Therefore, the reciprocals are in HP
a/(b+c),b/(a+c),c/(a+b) are in HP.