How do I find the sum of a sequence whose common difference is in Arithmetic Progression? How do I find the sum of a sequence whose common difference is in Arithmetic Progression ?
Like in the following series :-
$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$
And also how to find it's $n^{th}$ term ??

$$
\begin{align}
& \text{Let } a = a_{0} \space\,,\space d = \frac{1}{n} (a_{n} - a_{n-1}) \space\Rightarrow\space a_{n} = a_{n-1} + d \cdot n \quad \colon n \ge 1 \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow a_{n} = a_{n-1} + n = \{ \text{1, 3, 6, 10, ...} \} } \\
& \\
& a_{n} = d (n) + a_{n-1} = d (n) + d (n-1) + a_{n-2} = d (n) + d (n-1) + \text{...} + d (2) + d (1) + a \Rightarrow \\
& a_{n} = a + d \sum_{k=1}^{n} (k) = a + \frac{d}{2} n (n+1) \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow a_{n} = \frac{1}{2} n(n+1) = \{ \text{1, 3, 6, 10, ...} \} } \\
& \\
& \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} \left( a + \frac{d}{2} k(k+1) \right) = a \cdot n + \frac{d}{2} \left[ \sum_{k=1}^{n} k + \sum_{k=1}^{n} k^2 \right] \Rightarrow \\
& \sum_{k=1}^{n} a_{k} = a n + \frac{d}{2} \left[ \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} \right] = a n + \frac{d}{6} n(n+1)(n+2) \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow \sum_{k=1}^{n} a_{k} = \frac{1}{6} n(n+1)(n+2) = \{ \text{1, 4, 10, 20, ...} \} }
\end{align}
$$
