Question 1177330

we need to find n-th term formula, so first find differences

{{{6}}}........{{{6}}}........{{{2}}}........{{{-6}}}........{{{-18}}}
....{{{0}}}........{{{4}}}........{{{-8}}}........{{{-12}}}...........
.........{{{-4}}}.......{{{-4}}}........{{{-4}}}........................

if we look at the second difference they go up in steps of {{{-4 }}}
the next number in second row is {{{-12+(-4)=-16}}}
the next number in first row is {{{-18+(-16)=-34}}}

since second differences are equal, we conclude this  is a quadratic sequence

n-th term formula is:

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

to find {{{a }}}divide second difference by {{{2}}}

{{{a=-4/2=-2}}}

now we have

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

substitute {{{n=1}}} and {{{n=2}}} in formula above to get formulas for first two terms 

{{{a[1]=-2*1^2+b*1+c}}}
{{{a[2]=-2*2^2+b*2+c}}}

so, we have

{{{a[1]=-2+b+c}}}
{{{a[2]=-8+2b+c}}}

use first two terms

{{{6=-2+b+c}}} ............1)
{{{6=-8+2b+c}}}............2)
----------------------subtract 1) from 2)
{{{6-6=-8+2b+c -(-2+b+c )}}}
{{{0=-8+2b+c+2-b-c }}}
{{{0=-6+b}}}
{{{b=6}}}

go to

{{{6=-2+b+c}}} ............1), substitute {{{b}}}
{{{6=-2+6+c  }}}
{{{6=4+c}}}  
{{{c=6-4}}}
{{{c=2}}}

and your formula is:

{{{a[n]=-2n^2+6n+2}}}

now check if  {{{-6838}}} is in the sequence

{{{-6838=-2n^2+6n+2}}} ...........solve for {{{n}}}, the number of that term

{{{2n^2 - 6n - 6840 = 0}}} ...........simplify
 
{{{n^2 - 3n - 3420 = 0}}} ........using quadratic formula {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} we get

{{{n=-57}}} or{{{ n=60}}}

disregard negative solution, so

{{{n=60}}} and {{{60}}}th term is {{{-6838}}}