1)a, b, c are in AP

2)a, 2b, 3c are in AP

3)a, b, c are the sides of the triangle

4)none of the above

A)

log a, log b ,log c are in AP loga + logc = 2logb logac = logb^2 ac = b^2 .....................so a,b,c are in GP. log a-log 2b ,log2b-log3c , log3c-loga are in AP loga - log2b + log3c - loga = 2 (log2b - log3c) log(3c/2b) = 2 (log (2b/3c)) 3c/2b = (2b/3c)^2 THUS, 3c/2b = 1 3c = 2b b = 3c/2 ac= b^2 = 9c^2/4 a = 9c/4 b = 3c/2 c = c since this is a pythagorean triplet (sum of any 2 is greater than 3rd) So they are the sides of a triangle. |

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