IcE CoLd PrObLeM

The difference of this series is in arithmetic progression....i.e. 4,7,10 nd so on....so first we can find the sum of this a.p. upto 1st to 9999th term nd then we have to add this sum to the first term of our original sequence....and we will get 10000th term of the series... S=(n/2)(2a+(n-1)d)=(9999/2)(8+(9998)3)=149994999 Add this to the first term which is 1...so 149995000....which is the 10000th term...

In such cases when the differences are in AP, the $n^{th}$ term is defined as $t_{n} = an^{2} + bn + c$ .

Substituting n as 1,2,3 we can find a,b and c.

$t_{n} = \frac{3}{2} n^{2} - \frac{1}{2}n$

Substitute $n= 10000$ to get the answer as $149995000$