Answer
Arithmetic;
$71.5$
Work Step by Step
Let's note by $\{a_n\}$ the given sequence. The sum of its first $k$ terms is:
$$S_k=3+3.7+4.4+\dots+10.$$
The terms of the sequence are:
$$\begin{align*}
a_1&=3\\
a_2&=3.7\\
a_3&=4.4\\
&\dots\\
a_k&=10.
\end{align*}$$
We notice that the common difference of consecutive terms is $3.7-3=4.4-3.7=0.7$, therefore constant, so the sequence is arithmetic with the elements:
$$\begin{cases}
a_1=3\\
d=0.7.
\end{cases}$$
Before calculating $S_k$ we must find $k$ (the number of terms in the sum):
$$\begin{align*}
a_k&=a_1+(k-1)d\\
10&=3+(k-1)(0.7)\\
10&=3+0.7k-0.7\\
7.7&=0.7k\\
k&=11.
\end{align*}$$
Calculate the partial sum $S_{11}$ using the formula:
$$S_n=\dfrac{n(a_1+a_n)}{2}.$$
For $n=11$ we have:
$$S_{11}=\dfrac{11(3+10)}{2}=71.5.$$
