(frustum BF): (circumscribed figure) 
= n {a. nh + (nh) 2 } : {a. rh + (rh) 2 }. 
But, by Prop. 2, 
n[a. nh + (nh) 2 } : Sj” -1 {a. rh + (rh) 2 } > (a + nh):{^a + ^nh) 
> n{a. nh + (nh) 2 } : 2 t w {a . rh + (rh) 2 }. 
From these relations it is inferred that 
(frustum BF): (volume of segment) = (a + nh): (^a + ^nh), 
(volume of segment): (volume of cone ABB') 
- (AD+ 3CA): (AD +.2CA); 
and this is confirmed by the method of exhaustion. 
The result obtained by Archimedes is equivalent to proving 
that, if h be indefinitely diminished while n is indefinitely 
increased but nh remains always equal to b, then 
limit of n (ah + h 2 ) / S n = (a + h) / (-| a + h), 
limit of - S n = h 2 (-|a + §b), 
S n = a (h + 2h + 3h + ... + nh) + [K 2 + (2h) 2 + (3h) 2 + ... + (nh) 2 } 
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