Question 1161258

 {{{S}}} is an arithmetic series.

so, nth term formula is:

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

if {{{S[12] = 150}}}, we have

{{{150=S[1]+d(12-1)}}}

{{{150=S[1]+11d}}}.......solve for {{{S[1]}}}

{{{S[1]=150-11d}}}......eq.1


if {{{S[25] = 875}}}, we have

{{{875=S[1]+d(25-1)}}}

{{{875=S[1]+24d}}}.....solve for {{{S[1]}}}

{{{S[1]=875-24d}}}......eq.2

from eq.1 and eq.2 we have

{{{150-11d=875-24d}}}...solve for {{{d}}}

{{{24d-11d=875-150}}}

{{{13d=725}}}

{{{d=725/13}}}

go to

{{{S[1]=150-11d}}}......eq.1, substitute {{{d}}}

{{{S[1]=150-11(725/13)}}}

{{{S[1]=150-7975/13}}}

{{{S[1]=-6025/13}}}


your formula is:

{{{S[n]=-6025/13+(725/13)(n-1)}}}


then,  the value of {{{S[40]}}} is:

{{{S[40]=-6025/13+(725/13)(40-1)}}}

{{{S[40]=-6025/13+(725/13)(39)}}}

{{{S[40]=-6025/13+2175}}}

{{{S[40]=22250/13}}}

{{{S[40]=1711.5384615384617}}}....round to {{{2 }}}decimal places

{{{S[40]=1711.54}}}