Question 1135126
The formula for nth term of an arithmetic progression is 

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

In this example we have{{{ d=-5}}} and{{{ a[1]=1}}}

{{{a[n]=1+(n-1)(-5)}}}

{{{a[n]=1-5n+5}}}

{{{a[n]=-5n+6}}}

first  the sum of the first {{{12}}} terms 
{{{n}}} | {{{6 - 5 n}}}
{{{1}}} | {{{1}}}
{{{2 }}}|{{{ -4}}}
{{{3}}} | {{{-9}}}
{{{4}}} | {{{-14}}}
{{{5 }}}| {{{-19}}}
{{{6}}} |{{{ -24}}}
{{{7}}} | {{{-29}}}
{{{8 }}}| {{{-34}}}
{{{9}}} | {{{-39}}}
{{{10}}} | {{{-44}}}
{{{11}}}|{{{-49}}}
{{{12}}}|{{{-54}}}


{{{sum=1-4-9-14-19-24-29-34-39-44-49-54= -318}}}