Question 817762
<pre>
There are 9 arithmetic means between 11 and 51. The sum of the progression is

11, _, _, _, _, _, _, _, _, _, 51

Notice with 9 terms between them, that 
gives a sequence of 11 terms, counting 
the 9 and the 51, so n=11, a<sub>1</sub>=11,
a<sub>11</sub> = 51.  (There are two elevens, 
so don't get them mixed up :). 

If we use this sum formula, we don't need d:

S<sub>n</sub> = {{{n/2}}}(a<sub>1</sub> + a<sub>n</sub>)

S<sub>11</sub> = {{{11/2}}}(11 + 51)

S<sub>11</sub> = {{{11/2}}}(62)

S<sub>11</sub> = 341


Edwin</pre>