Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
if a^2(b+c), b^2(c+a), c^2(a+b) are in A.P, prove that either a,b,c are in A.P or ab+bc+ca=0
Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, b²(c+a) - a²(b+c) = c²(a+b) - b²(c+a)==> b²c + b²a - a²b - a²c = c²a + c²b - b²c - b²a==>(b²c - a²c) + (b²a - a²b) = (c²a - b²a) + (c²b - b²c)==>c(b² - a²) + ab(b-a) = a(c² - b²) + bc(c - b)==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )∴ Either : ab + bc + ca = 0Or : b - a = c - b, i.e., a, b, c are in A.P.Thanks, Vj
If are in AP adding, abc to each term, are also in AP i.e. are in AP Now, if ab+bc+ca is non-zero, then dividing by ab+bc+ca, we get a,b,c are in AP
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -