Finr the first term and the difference of an arithmetic progression, given two relations between its terms In an AP, the 15th term is double the 9th term. If also, the sum of the first 15 and the sum of the first 9 terms added together is 279, find the first term and the common difference.


Alright, so we have to make two equations and solve them simultaneously. 

First equation is pretty easy: 

$$a + 14d = 2(a+8d)  $$

How do we make the second equation? the formula for sum of an AP is:  

$S_n = (\frac n2)[2a + (n-1)d]$ where a is the first term and d is the common difference

We can't use the value of 279 because  that's for the 15th and the 9th term.
The sum of the first $15$ terms is $\frac{15}2 (2a + 14d)$, and the sum of the first $9$ terms is $\frac{9}{2}(2 a + 8d)$.

Their sum together is $279$, so
$$\frac{15}2 (2a + 14d) + \frac{9}{2}(2 a + 8d) = 279$$

is your second equation.
Each new term equals the previous term plus the common difference. So for 9 terms we add the common difference 8 additional times for a total of sum of 1 to 8 with the common difference of 1, which is 36. Hence the second equation:
         9a+36d=279 or a+4d=31
solve with the first equation of
         2(a+8d)=a+14d or a=-2d
The solution is:
  a=-31
  d=15.5