Question:
Find the A.M. between:
(i) 7 and 13
(ii) 12 and −8
(iii) (x − y) and (x + y).
Solution:
(i) Let $A_{1}$ be the A.M. between 7 and $13 .$
$A_{1}=\frac{a+b}{2}$
$=\frac{7+13}{2}$
= 10
(ii) Let $A_{1}$ be the A.M. between 12 and $-8$.
$A_{1}=\frac{a+b}{2}$
$=\frac{12+(-8)}{2}$
=2
(iii) Let $A_{1}$ be the A.M. between $(x-y)$ and $(x+y)$.
$A_{1}=\frac{a+b}{2}$
$=\frac{(x-y)+(x+y)}{2}$
$=x$
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