2012/015) find Nth term of this sequence? 14, 29, 54, 89, 134

The 1st order differences are 15,25,35,45

The second order differences are 10, 10,10,

So the equation for the 1st order differences are 5 + 10n

Now the original terms are equation is 2nd order that is an^2+bn+c

difference is

a(n+1)^2 + b(n+1) + c – an^2 –bn – c
= a(2n+1) + b = 2an + (a+b) = 10n + 5
So a = 5 and b= 0

So we have nth term = 5n^2+ c

for n=1 gives 5n^2 + c = 14 or c= 9

so nth term = 5n^2+9