This is an arithetic sequence because the difference between each two consecutive numbers is 4.
The formula for any term in an arithmetic sequence is: tn = t1 + (n - 1)d
tn = t35 (the 35th term) t1 = 3 (the first term) n = 35 (the 35th term)
d = 4 (the common difference)
tn = t1 + (n - 1)d ---> t35 = 3 + (35 - 1)·4 ---> t35 = 139
35th term of Arithmetic sequence 3,7,11,15,19
\(\small{\text{We have $t_x=t_1=3$ and $t_y=t_{2}=7$ and we want $t_z=t_{35}=?$ }}\\\\ \small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right) + t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~} $}}\\\\ \small{\text{$ \begin{array}{rcl} t_{35} &=& t_1\cdot \left( \dfrac{2-35}{2-1}\right) + t_{2}\cdot \left(\dfrac{35-1}{2-1}\right) \\\\ t_{35} &=& 3\cdot \left( \dfrac{-33}{1}\right) + 7\cdot \left(\dfrac{34}{1}\right) \\\\ t_{35} &=& -3\cdot 33 + 7\cdot 34 \\ t_{35} &=& -99 + 238 \\ \mathbf{t_{35}} & \mathbf{=} & \mathbf{139}\\ \hline \\ \end{array} $}}\)