respected sir,
Richard is a harsh teacher.He gave a homework:
(b-c)^2 , (c-a)^2, & (a-b)^2 , are in arithmatical progression,--- if this be true then prove:
1/(b-c) , 1/(c-a), 1/(a-b) are in AP too.
Sam a cunning guy di d the problem in the following way:
if the series of squares be in AP then the following is also in AP:
(b-c)^2 - ( a^2+ b^2+ c^2 -ab -bc -ca)
(c-a)^2 - (a^2+ b^2+ c^2 - ab-bc-ca)
(a-b)^2 - (a^2+ b^2+ c^2 -ab-bc-ca)--------------are in AP
i.e., (a-b)(c-a) ,(b-c)(a-b) ,(b-c)(c-a) are in AP
dividing by (a-b)(b-c)(c-a) we get the proof
on showing the H.W., Richard asked Sam " How could one think out intuitively
that (a^2+b^2+c^2-ab-bc-ca) need to be subsracted & how to get the hint from the problem"?
WHAT WAS SAM'S ANSWER? -------please explain lucidly :?
Richard is a harsh teacher.He gave a homework:
(b-c)^2 , (c-a)^2, & (a-b)^2 , are in arithmatical progression,--- if this be true then prove:
1/(b-c) , 1/(c-a), 1/(a-b) are in AP too.
Sam a cunning guy di d the problem in the following way:
if the series of squares be in AP then the following is also in AP:
(b-c)^2 - ( a^2+ b^2+ c^2 -ab -bc -ca)
(c-a)^2 - (a^2+ b^2+ c^2 - ab-bc-ca)
(a-b)^2 - (a^2+ b^2+ c^2 -ab-bc-ca)--------------are in AP
i.e., (a-b)(c-a) ,(b-c)(a-b) ,(b-c)(c-a) are in AP
dividing by (a-b)(b-c)(c-a) we get the proof
on showing the H.W., Richard asked Sam " How could one think out intuitively
that (a^2+b^2+c^2-ab-bc-ca) need to be subsracted & how to get the hint from the problem"?
WHAT WAS SAM'S ANSWER? -------please explain lucidly :?