Summing $3+7+14+24+37...$ up to $10$ terms 
What is $3+7+14+24+37...$ up to $10$ terms? 

I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?

This answer seeks to expand on the "one can check that" comment in Foobaz's answer. First, let $A=\{3,7,14,24,37,\ldots\}$. The pattern for the difference, $d_n$, between terms $n$ and $n+1$ is clear enough:
$$
d_n=4+(n-1)3=1+3n.
$$
Hence,
$$
a_{n+1}=\frac{n}{2}(5+3n)+3
$$
or, more usefully,
\begin{align}
a_n&=\frac{n-1}{2}(2+3n)+3\\[1em]
&=(n-1)+\frac{3n^2-3n}{2}+3\\[1em]
&= \frac{2n-2+3n^2-3n+6}{2}\\[1em]
&= \frac{3n^2-n+4}{2}.
\end{align}
Thus, we have the following:
\begin{align}
S_{10}&= \sum_{i=1}^{10}\frac{3i^2-i+4}{2}\\[1em]
&= \frac{3}{2}\sum_{i=1}^{10}i^2-\frac{1}{2}\sum_{i=1}^{10}i+2\sum_{i=1}^{10}1\\[1em]
&= \frac{3}{2}\left[\frac{10(10+1)(2(10)+1)}{6}\right]-\frac{1}{2}\left[\frac{10(10+1)}{2}\right]+2(10)\\[1em]
&= 570.
\end{align}
