Question 877961
Formula for Arithmetic Series, {{{S=(n/2)(a[1]+a[n])}}}  or {{{S=(n/2)(2a[1]+(n-1)d)}}}  in which d is the common difference between sequence terms.


Your equation does not indicate the actual last term or its index.  The formula using the common difference should be a good choice.


Your equation shows d=5 and {{{a[1]=1}}}.
{{{(n/2)(2*1+(n-1)*5)=148}}}, applying that second version of the formula.
{{{(n/2)(2+5n-5)=148}}}
{{{n(-3+5n)=296}}}
{{{5n^2-3n=296}}}
{{{highlight_green(5n^2-3n-296=0)}}}.
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Discriminant:  9+4*5*296=5929, and sqrt(5929)=77.
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GENERAL SOLUTION FOR QUADRATIC EQUATION GIVES THIS RESULT...
{{{n=(3+- 77)/(2*5)}}}, the PLUS-form makes sense.
{{{n=(3+77)/10}}}
{{{n=80/10}}}
{{{highlight(n=8)}}}
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So what is x ?
x is the eighth term of the sequence.
General term of the sequence:  {{{1+(n-1)*5}}};
using n=8, {{{x=1+(8-1)*5}}},
{{{x=1+7*5}}}
{{{highlight(highlight(x=36))}}}.
You can fill-in the other terms if you want.



1+6+11+16+21+26+31+36=148