SamRoy's answer is correct.
Here is a demonstration of how to compute the answer in practice
If the first term of the arithmetic sequence is $a$ and the common difference is $d$ then the five terms are $$a, a+d,a+2d, a+3d, a+4d$$ Similarly, if the first term of the geometric sequence is $b$ and the common ratio is $r$ then the terms in the geometric sequence are $$ b, br, br^2, br^3, br^4$$ So we have $$ a+b = 131 \Rightarrow a = 131-b$$ $$ a+d+br = 157 \Rightarrow (131-b)+d+br = 157 \Rightarrow d = 26+(1-r)b$$ $$ a+2d+br^2 = 211 \Rightarrow (131-b) + 2(26+(1-r)b) + br^2 = 211$$ $$ \Rightarrow 183+b-2br +br^2 = 211 \Rightarrow b = \frac{28}{(r-1)^2} \Rightarrow d = 26 - \frac{28}{r-1} $$ $$ \Rightarrow a = 131 - \frac{28}{(r-1)^2}$$ Finally, we have $$ a+3d + br^3 = 349 \Rightarrow \left(131 - \frac{28}{(r-1)^2}\right) + 3\left(26 - \frac{28}{r-1}\right) + \frac{28r^3}{(r-1)^2} = 349$$ $$ \Rightarrow \frac{28r^3 -84r+56}{(r-1)^2} = 140 \Rightarrow 28\left(\frac{(r-1)^2(r+2)}{(r-1)^2}\right) = 140$$ and because the expressions for $a$, $b$ and $d$ allow us to conclude that $r \neq 1$, we can divide above and below by $(r-1)^2$ to get $$ 28(r+2) = 140 \Rightarrow r = 3$$ From the above expressions, we then immediately get $$a = 124, b = 7, d = 12$$ and we just need to do the final check that $$ a + 4d + br^4 = 739 $$