0
$\begingroup$


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.

$\endgroup$
  • $\begingroup$ Here is something worth reading: en.wikipedia.org/wiki/Triangular_number $\endgroup$ – Colm Bhandal Jul 25 '15 at 16:22
  • 3
    $\begingroup$ Arithmetic Progression with dynamic common difference is not arithmetic progression ;) $\endgroup$ – Michael Galuza Jul 25 '15 at 16:24
  • $\begingroup$ Thanks a lot this wiki really helped me out.. $\endgroup$ – Palash Taneja Jul 26 '15 at 3:19
0
$\begingroup$

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 $=1$

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

$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.