Sum the array times n, except the last I've been posting relatively hard challenges recently, so here goes an easy one.
Task
Given an array \$A\$ and a number \$n\$, calculate the sum of all numbers of \$A\$ multiplied by \$n\$, except the last one. All numbers (the elements of \$A\$ and the value of \$n\$) are positive integers, and \$A\$ is non-empty. Shortest code in bytes wins.
I have a 3-byte J solution. Can you find it (or beat it in a different language)?

Test cases
A           N   Ans   Explanation
3 1 4 1 5   10  95    (3+1+4+1)*10+5
3 1 4 1 5   1   14    (3+1+4+1)*1+5
1           999 1     1
J, 3 bytes
That was fun to find.
&amp;+/

Try it online!
How it works
10 (&amp;+/) 3 1 4 1 5 will bind 10 as an argument of + as 10&amp;+, one verb that gets inserted between the elements of the list by /. So we have: 3 (10&amp;+) 1 (10&amp;+) 4 (10&amp;+) 1 (10&amp;+) 5. Now x n&amp;v y means that y gets applied to n&amp;v for x times. With J's right to left evaluation we get: to 5 add 1 times 10, add 4 times 10, add 1 times 10, add 3 times 10. A challenge made for J's stranger parts. :-) And because + is commutative, +&amp;/ would also be a valid solution.