If arithmetic mean and geometric mean of two positive numbers a and b are 7 and √6 respectively, then the value of (a3-15 a2+20 a-5) is equal to

Question

If arithmetic mean and geometric mean of two positive numbers a and b are 7 and √6 respectively, then the value of (a3-15 a2+20 a-5) is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

( a + b /2)=7, ab =6 ⇒ b =(6/ a ) ⇒ a 2+6-14 a =0 ⇒ a ( a 2-14 a +6)-1( a 2-14 a +6)+1 ⇒ 1

Q. If arithmetic mean and geometric mean of two positive numbers $a$ and $b$ are 7 and $6$ respectively, then the value of $(a^{3} - 15 a^{2} + 20 a - 5)$ is equal to

1

2

3

4

$\frac{a + b}{2} = 7, ab = 6$
$\Rightarrow b = \frac{6}{a} \Rightarrow a^{2} + 6 - 14 a = 0$
$\Rightarrow a (a^{2} - 14 a + 6) - 1 (a^{2} - 14 a + 6) + 1$
$\Rightarrow 1$

Q. If arithmetic mean and geometric mean of two positive numbers a and b are 7 and 6​ respectively, then the value of (a3−15a2+20a−5) is equal to

1

2

3

4

2a+b​=7,ab=6 ⇒b=a6​⇒a2+6−14a=0 ⇒a(a2−14a+6)−1(a2−14a+6)+1 ⇒1

Q. If arithmetic mean and geometric mean of two positive numbers $a$ and $b$ are 7 and $6$ respectively, then the value of $(a^{3} - 15 a^{2} + 20 a - 5)$ is equal to

$\frac{a + b}{2} = 7, ab = 6$
$\Rightarrow b = \frac{6}{a} \Rightarrow a^{2} + 6 - 14 a = 0$
$\Rightarrow a (a^{2} - 14 a + 6) - 1 (a^{2} - 14 a + 6) + 1$
$\Rightarrow 1$