Question

STATEMENT-1 : If $a{b}^2{c}^3,a^2{b}^3{c}^4,a^3{b}^4{c}^5$ are in A.P.
$(a,b,c>0),$ then the minimum value of a + b + c is 3.

STATEMENT-2 : Arithmetic mean of any two numbers is greater than geometric mean of the numbers.

• A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is the correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

Answer: (C)

STATEMENT-1 is True, STATEMENT-2 is False

$\frac{a+b+c}{3}\ge (abc{)}^{1/3}\Rightarrow a+b+c\ge 3(abc{)}^{1/3}$
$\Rightarrow 1,abc,a^2{b}^2{c}^{2}inA.P.\because abc\ne 0$
$2abc=1+a^2{b}^2{c}^2$
$\Rightarrow abc=1soa+b+c\ge 3$
Hence minimum values of $a+b+c$ is 3.
Consider the numbers $-2$ and $8$. Their $A.M.=-5$ and G.M.
$=\sqrt{(-2)(-8)}=4$
$\therefore A.M.<G.M.$
$\therefore$ Statement - 2 is false.