Question 928076
I assume {{{a}}}= first term, {{{t[10]}}}= tenth term, {{{t[19]}}}= term number {{{19}}} , and {{{S[23]}}}= sum of first {{{23}}} terms.
It is commonly understood that {{{d}}}= common difference.
{{{t[n]=a+(n-1)d}}} , so
{{{system(21=a+9d,48=a+18d)}}} --->{{{48-21=18d-9d}}}--->{{{27=9d}}}--->{{{27/3=d}}}--->{{{highlight(d=3)}}}
Then, {{{system(21=a+9d,d=3)}}} --->{{{21=a+9*3}}}--->{{{21=a+27}}}--->{{{21-27=a}}}--->{{{highlight(a=-6)}}} .
There are different formulas to calculate sums of arithmetic sequences, but I prefer to remember that adding up the first {{{n}}} terms twice, grouping them in pairs "head-to-tail", {{{t[0]+t[n]}}} , {{{t[1]+t[n-1]}}} , {{{t[2]+t[n-2]}}} , and so on, you end up with a sum of {{{n}}} pairs that all add up to {{{t[0]+t[n]}}}, so
{{{2S[n]=n(t[0]+t[n])}}}-->{{{S[n]=(n(t[0]+t[n]))/2}}} .
So, {{{t[23]=-6+22*3=-6+66=60}}} and {{{S[23]=(23(a+t[23]))/2=(23(-6+60))/2=(23(54))/2=highlight(621)}}} .