The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 3/2 and the geometric mean exceeds their harmonic mean by 6/5. If…

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The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 3/2 and the geometric mean exceeds their harmonic mean by 6/5. If a+b=α and a-b=β, then the value of (10β)/(α) is equal to

MARKS Content Team · Accepted Answer

Correct answer: 6. A-G=3/2 and G-H=6/5 Also, we know that G^2=AH ⇒G^2=3/2+GG-6/5 ⇒G^2=G^2-6/5G+3/2G-9/5 ⇒3/2-6/5G=9/5⇒G=9/5×10/3=6 G=6⇒A=15/2 ⇒(a+b)/2=15/2 and √ab=6 ⇒a+b=15 and ab=36 a-b=√(a+b^2-4ab)=√(225-4×36)=9 α=a+b=15, β=a-b=9 ⇒(10β)/(α)=(10×9)/15=6

The arithmetic mean of two positive numbers $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean exceeds their harmonic mean by $\frac{6}{5} .$ If $a + b = α$ and $|a - b| = β,$ then the value of $\frac{10 β}{α}$ is equal to

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The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 32 and the geometric mean exceeds their harmonic mean by 65. If a+b=α and a-b=β, then the value of 10βα is equal to

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The arithmetic mean of two positive numbers $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean exceeds their harmonic mean by $\frac{6}{5} .$ If $a + b = α$ and $|a - b| = β,$ then the value of $\frac{10 β}{α}$ is equal to