Arithmetic Progression with dynamic common difference

While preparing for my recent exams I noticed some interesting AP problems which I was not able understand. The problem was that the common difference was itself in arithmetic progression

For Example: 1,3,6,10,15...

You can thus clearly see that the difference is in arithmetic progression as well

I would be grateful to anyone who could give me a solution to my issue.

Answer

Let $S_n=1+3+6+10+15+\cdots+T_{n-1}+T_n$ where $T_m$ is the $m(>0)$th term

$\ \ \ \ \ \ S_n=\ \ \ \ \ \ 1+3+6+10+15+\cdots+T_{n-1}+T_n$

On subtraction, $0=1+(3-1)+(6-3)+(10-6)+\cdots+(T_n-T_{n-1})-T_n$

$\implies T_n=1+2+3+\cdots$ up to $n$ terms which is clearly a de facto arithemetic Series with the first term $=1$ and common difference $=1$

$\implies T_n=\dfrac n2\{2\cdots1+(n-1)1\}=\dfrac{n(n+1)}2$