Question 194494
The multiples of 6 form an arithmetic sequence with common 
difference, d = 6.
<pre><font size = 4 color = "indigo"><b>
3138 and 51804 are both multiples of 6.  However the fact that

it says "strictly between 3138 and 51804", we must begin with

the first multiple of 6 larger than than 3138, which is 3144.

Also we must end with largest multiple of 6 less than 51804,

which is 51798.  So,

{{{a[1]=3144}}}, {{{a[n]=51798}}}

The formula for the nth term is

{{{a[n]=a[1]+(n-1)d}}}

{{{51798=3144+(n-1)6}}}

{{{51798=3144+6(n-1)}}}

{{{51798=3144+6n-6}}}

{{{51798=3138+6n}}}

{{{51798=3138+6n}}}

{{{48660=6n}}}

{{{8110=n}}}

The formula for the sum of the n terms 
of an arithmetic sequence is

{{{S[n]=n(a[1]+a[n])/2}}}  

{{{S[8110]=8110(3144+51798)/2}}}

{{{S[8110]=8110(54942)/2}}}

{{{S[8110]=8110(54942)/2}}}

{{{S[8110]=222789810}}}

Edwin</pre>