Question

the arithmetic mean of two numbers exceed their harmonic mean by 27.The lowest of the two numbers is​

Answer 1

Step-by-step explanation:

Let the numbers be a, b and their A.M., G.M., and H.M. be denoted by

A,G, and H respectively. Also we know that A.G.H are in G.P.

or G

2

=AH ...(1)

Since A−G=15 and A−H=27

(A−15)

2

=G

2

=AH by (1) =A(A-27)

or −30A+225=−27Aor3A=225

∴A=75=

2

a+b

∴a+b=150 ...(2)

Since A−G=15∴75−G=15

or G=60=

ab

ab=3600...(3)

Hence from (2) and (3) we conclude that a and b are the roots of t

2

−150t+3600=0

or (t−120)(t−30)=0∴t=120,30

Hence the two numbers are 120 and 30