If an arithmetic progression starts with 4, what is the common difference if the sum of the first 12 terms is twice the sum of the first 8 terms? An arithmetic progression (AP) has 4 as its first term. What is the common difference if the sum of the first 12 terms is 2 times the sum of the first 8 terms?
 A: Hint: Let $d$ be the common difference. The first $8$ terms are $4,4+d,4+2d,...,4+7d.$ Do you know how to find their sum? (If you aren't sure how to add up the terms with the $d$s, you might look up the triangular numbers.) The first $12$ terms are $4,4+d,4+2d,...,4+11d.$ The same approach will allow you to find that sum.
What does the relationship between the two sums need to be? Solve the resulting equation for $d$.
A: Let $d$ be the common difference.  The sum of the first $12$ terms is then $$4 + (4 + d) + (4 + 2d) + \ ... \ + (4 + 11d)$$
Recall now the key fact that the sum of the first $n$ integers is $\frac{(n+1)n}{2}$.  Obviously you can write the expression for the sum of the first $8$ terms similarly and it should now reduce just to arithmetic.
A: HINT:
If $4$ is the first term and $d$ is common difference, the sum of first $n$ terms is $\frac n2\cdot \{2\cdot4+(n-1)d\}$ 
So, we have $$\frac{\frac {12}2\cdot \{2\cdot4+(12-1)d\}}{\frac 82\cdot \{2\cdot4+(8-1)d\}}=\frac21$$
A: Another way of looking at this, which is less general - the sum of the first eight terms is equal to the sum of the next four (this works because the multiple in the question is $2$, which makes things easy) so:$$4+(4+d)+(4+2d)+\dots (4+7d)=(4+8d)+(4+9d)+(4+10d)+(4+11d)$$
I reckon this is simpler arithmetic - lots can be easily cancelled.
