How to find nth term of the sequence 6+13+24+39+.....
Hey roshanhero.
For your problem without any more clarification there will be an infinite number of sequences that fit your description.
You could find one by fitting a polynomial through them or use a program to generate a sequence.
Without any more constraints, a unique sequence is not really possible.
I would do this using "Newton's divided difference formula". The numbers are $\displaystyle a_0= 6$, $\displaystyle a_1= 13$, $\displaystyle a_2= 24$, and $\displaystyle a_3= 39$. The "First Differences" are $\displaystyle \Delta a_0= 13- 6= 7$, $\displaystyle \Delta a_1= 24- 13= 11$, $\displaystyle \Delta a_2= 39- 24= 15$ and then the "Second Differences" are $\displaystyle \Delta^2 a_0= 11- 7= 4$, $\displaystyle \Delta^2 a_1= 15- 11= 4$. If we assume that the sequence continues so that the second differences are alway four (and the third and higher differences are 0) we can write, in analogy with the Taylor's series, $\displaystyle a_n= a_0+ \Delta a_0 n+ (\Delta^2 a_0/2)n(n-1)= 6+ 7n+ (4/2)n(n- 1)= 6+ 7n+ 2n^2- 2n= 6+ 5n+ 2n^2$.
You will note that when n= 0, $\displaystyle a_0= 6+ 5(0)+ 2(0^2)= 6$, when n= 1, $\displaystyle a_1= 6+ 5(1)+ 2(1^2)= 13$, when n= 2, $\displaystyle a_2= 6+ 5(2)+ 2(2^2)= 24$, and when n= 3, $\displaystyle a_3= 6+ 5(3)+ 2(3^2)= 39$ exactly as we want.
You could also use the fact that, since the second differences are constant, the sequence is a quadratic function of n: $\displaystyle a_n= an^2+ bn+ c$. Then use $\displaystyle a_0= a(0)+ b(0)+ c= c= 6$,
$\displaystyle a_1= a(1^2)+ b(1)+ c= a+ b+ c= 13$ and $\displaystyle a_2= a(2^2)+ b(2)c= 4a+ 2b+ c= 24$ to solve for a, b, and c.