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Question:

For any three positive real numbers a, b, and c, 9(25a²+b²)+25(c²-ac)=15b(3a+c), then.

b,c and a are in G.P.

a, b and c are in A.P.

a, b and c are in G.P

b, c and a are in A.P.

Solution:

9(25a²+b²)+25(c²-ac)=15b(3a+c)
225a²+9b²+25c²-25ac=45ab+15bc
225a²+9b²+25c²-25ac-45ab-15bc=0
(15a)²+(2b)²+(5c)²-(15a)(5c)-(15a)(3b)-(3b)(5c)=0
(15a)²+(2b)²+(5c)²=(15a)(5c)+(15a)(3b)+(3b)(5c)
15a=3b=5c=k
b=5a and c=3a
Hence, a=k/15, b=k/3 and c=k/5
∴b-a=4k/15 and c-a=2k/15 and b-c=2k/15
∴c-a=b-c ⇒ 2c=a+b
Hence, b, c and a are in A.P.