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.

• Here is something worth reading: en.wikipedia.org/wiki/Triangular_number – Colm Bhandal Jul 25 '15 at 16:22
• Arithmetic Progression with dynamic common difference is not arithmetic progression ;) – Michael Galuza Jul 25 '15 at 16:24
• Thanks a lot this wiki really helped me out.. – Palash Taneja Jul 26 '15 at 3:19

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 $=1$

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

The sequence you listed is the sequence of triangular numbers. The $n^{th}$ triangular number, $T_n$, is of the form $$T_n = \frac{n(n+1)}{2}$$

• Does this formula hold good for other similar sequences like this one: 1,4,8,13,19,26... – Palash Taneja Jul 26 '15 at 3:20
• No. It holds only for the triangular numbers. You may be able to derive similar formulas for other arithmetic progressions. – Race Bannon Jul 26 '15 at 3:24