Question 1197256


Find the {{{10}}}th term of the sequence 

{{{-37}}},{{{-34}}},{{{-41}}}


nth term equation is:

{{{a[n]=an^2+bn+c}}}


first term: {{{-37}}} =>{{{n=1}}}
second term:{{{ -34}}} =>{{{n=2}}}
third term: {{{-41 }}}=>{{{n=3}}}


make system of equations:

 {{{ -37=a*1^2+b*1+c}}}
{{{-34=a*2^2+b*2+c}}}
{{{-41=a*3^2+b*3+c}}}
_____________________

{{{  -37=a+b+c}}}..........eq.1
{{{-34=4a+2b+c}}}..........eq.2
{{{-41=9a+3b+c}}}..........eq.3
_____________________

subtract eq.1 from eq.2

{{{-34-(-37)=4a+2b+c-a-b-c}}}
{{{3=3a+b}}}
{{{b=3-3a}}}..........eq.1a


subtract eq.1 from eq.3

{{{-41-(-37)=9a+3b+c-a-b-c}}}
{{{-4=8a + 2b}}}.....divide by {{{2}}}
{{{-2=4a + b}}}
{{{b=-2-4a}}}.............eq.2a

from eq.1a and eq.2a we have

{{{3-3a=-2-4a}}}

{{{4a-3a=-2-3}}}

{{{a=-5}}}


go to

{{{b=3-3a}}}..........eq.1a, substitute {{{a}}}

{{{b=3-3(-5)}}}

{{{b=3+15}}}

{{{b=18}}}


go to


{{{  -37=a+b+c}}}..........eq.1, substitute {{{a}}} and {{{b}}}


{{{  -37=-5+18+c}}}..........solve for {{{c}}}

{{{  -37+5-18=c}}}

{{{c=-50}}}

so, solving this system we got

{{{a = -5}}}, {{{b = 18}}}, {{{c = -50}}}


then nth term equation is:

{{{a[n]=-5n^2+18n-50}}}


now we can find  the {{{10}}}th term of the sequence
 
{{{n=10}}}

{{{a[10]=-5*10^2+18*10-50}}}

 {{{a[10]=-370}}}