Arithmetic series has first term An arithmetic series has first term a and common difference d. 
The sum of the first 31 terms of the series is 310 
a) Show that a + 15d = 10 
b) Given also that the 21st term is twice the 16th term, find the values of a and d. 
 A: HINT: The first $31$ terms are $a,a+d,a+2d,\dots,a+30d$, so their sum is
$$\begin{align*}
\underbrace{a+(a+d)+(a+2d)+\ldots+(a+30d)}_{31\text{ terms}}&=31a+(d+2d+3d+\ldots+30d)\\
&=31a+d(1+2+3+\ldots+30)\;.
\end{align*}$$
Do you know the formula for the sum of the first $n$ positive integers? Using it (or any other legitimate method) calculate $1+2+\ldots+30$. Call this sum $s$. Then you’re told that $31a+ds=310$. If you’ve done it right, you’ll be able to simplify that equation to show that $a+15d$ is a particular number, but that number isn’t $0$: you have a typo in the problem.
For the second part, use the fact that the $16$-th term is $a+15d$ (why $15$ and not $16$?), and the $21$-st is $a+20d$. Express the fact that $a+20d$ is twice $a+15d$ as a linear equation in $a$ and $d$, combine it with the equation $31a+ds=310$ (or the simplified version that you got in part (a)), and solve the resulting system of two linear equations in the unknowns $a$ and $d$.
A: $$(31/2)(2a_1+30d)=310$$
$$(1)...a_1+15d=10$$
$$a_{21}=2a_{16}$$
$$a_1+20d=2a_1+30d$$
$$(2)...a_1=-10d$$
put $(2)$ on$(1)$ we get $d=2$ then get $a_1=-20$
A: a) We know that the nth partial sum of an arithmetic sequence is $$S_n = {n(a_1 + a_n)\over2}$$ and that $$a_n = a_1 + (n-1)d$$
since we know $S_{31} = 310 = {31\over2}(a_1+a_{31})$ and $a_{31} = a_1 + 30d$, we can substitute to get $$310 = {31\over2} (2a_1+30d) = 31(a_1+15d)$$
Dividing both sides by 31, we get$$a_1+15d = 10$$
b) Knowing that $a_{21} = 2a_{16}$ allows us to substitute those numbers into our second equation. $$a_1 + 20d = 2a_1 + 30d$$
simplified, $a_1 = -10d$.
Now we use that with the equation we proved in a) and substitute to get $5d=10$ or $d=2$
We then use $d$ to find $a_1=-20$.
