# If a, b, c are in A. P., a2, b2, c2 are in H. P. , then prove that either a = b = c or a, b, -c/2 form a G. P.

8 years ago
Hello Student,
Given that a, b, c are in A. P.
⇒ 2b = a + c ……. (1)
And a2, b2, c2 are in H. P.
$\frac{1}{b^2}-\frac{1}{d^2}=\frac{1}{c^2}-\frac{1}{b^2}$
⇒ (a – b)(a + b)/b2a2 = (b – c) (b + c)/b2c2
⇒ ac2 + bc2 = a2b + a2c [∵ a – b = b – c]
⇒ ac (c – a) + b (c – a) (c + a) = 0
⇒ (c – a) (ab + bc + ca) = 0
⇒ either c – a = 0 or ab + bc + ca = 0
⇒ either c = a or (a+ c) b + ca = 0 and then form (i) 2b2 + ca = 0
Either a = b = c or b2 = a (-c/2)
i.e. a, b, -c/2 are in G. P. Hence Proved

Thanks
$1/d^2 will be 1/a^2$ you have written wrong...as d is not a term given in the question ...pls look in the matter and coreect it