Question 1182229


given: {{{10}}},{{{30}}},{{{60}}},{{{100}}},{{{150}}},{{{210}}}

Sequence	:	{{{10}}}.......{{{30}}}.......{{{60}}}.......{{{100}}}.......{{{150}}}.......{{{210	}}} 		 		 		 		 	
First diff.	:	......{{{20}}}.......{{{30}}}.......{{{40}}}.......{{{50}}}.......{{{60}}}....... 		 		 		 		 		 
Second diff.	:	............{{{10}}}........{{{10}}}......{{{10}}}........{{{10}}}	

 		
We see that the second differences  are all equal so we concludet that this is a quadratic sequence.

The quadratic sequence has the form:
 
{{{T[n]=an^2+bn+c}}}

To find the value of  {{{a}}} we just divide second difference ( {{{10}}} ) with {{{2}}}.

{{{a=10/2=5}}}

Now we have:

{{{T[n]=5n^2+bn+c}}}
 

Substitute {{{n=1}}} and  {{{n=2}}} into above equation:

if {{{n=1}}}, {{{T[1]=10}}}

{{{10=5*1^2+b*1+c}}}

{{{10=5+b+c}}}

{{{10-5-b=c}}}

{{{5-b=c}}}....eq.1


if {{{n=2}}}, {{{T[2]=30}}}

{{{30=5*2^2+b*2+c}}}

{{{30=20+2b+c}}}

{{{30-20-2b=c}}}

{{{10-2b=c}}}.....eq.2

from eq.1 and eq.2 we have

{{{5-b=10-2b}}}

{{{2b-b=10-5}}}

{{{b=5}}}

go to

{{{5-b=c}}}....eq.1, substitute {{{b}}}

{{{5-5=c}}}

{{{c=0}}}

then nth term equation is:  {{{T[n]=5n^2+5n}}}