Question 957473
The {{{n}}} arithmetic means, {{{a[1]}}} , {{{a[2]}}} , {{{a[1]}}} , ...,{{{a[n]}}} ,
are equally spaced by a common difference {{{d}}} , forming an arithmetic sequence (or arithmetic progression):
{{{a[0]=75}}} ,
{{{a[1]=75-d}}} ,
{{{a[2]=75-2d}}} ,
{{{a[3]=75-3d}}} ,
{{{a[4]=75-4d}}} ,
................... ,
{{{a[n]=75-nd}}} , and
{{{19=75-(n+1)d}}} .
We do not know {{{n}}} , and we do not know {{{d}}} ,
but we have one equation relating them:
{{{19=75-(n+1)d}}} <---> {{{(n+1)d=75-19}}} <---> {{{(n+1)d=56}}} .
We just need another equation.
From the given ratio we need to get the other equation.
The easiest way I could think is to look at that arithmetic sequence/progression in reverse.
Starting from the other end
{{{a[n]=19+d}}} ,
{{{a[n-1]=19+2d}}} ,
{{{a[n-2]=19+3d}}} ,
{{{a[n-3]=19+4d}}} ,
{{{a[n-4]=19+5d}}} ,
and so on.
So, {{{a[n-3]+a[n-4]=(19+4d)+(19+5d)=38+9d}}} , and
we knew that {{{a[3]=75-3d}}} and {{{a[4]=75-4d}}} , so
{{{a[3]+a[4]=(75-3d)+(75-4d)=150-7d}}} .
The ratio of those sums is
{{{(150-7d)/(38+9d)=61/37}}}
Solving that equation for {{{d}}} we get
{{{(150-7d)/(38+9d)=61/37}}}
{{{37(150-7d)=61(38+9d)}}}
{{{37-150-37*7d=61*38+61*9d}}}
{{{5550-259d=2318+549d}}}
{{{5550-2318=259d+549d}}}
{{{3232=808d}}}
{{{3232/808=d}}} ---> {{{d=4}}}
Now, we plug that value for {{{d}}} into the equation
{{{(n+1)d=56}}} that we had, and get
{{{(n+1)*4=56}}} <--> {{{n+1=56/4}}} <--> {{{n+1=14}}} <--> {{{n=14-1}}} <--> {{{highlight(n=13)}}}