How can I get the first element in this sequence?

Question

There is an arithmetical sequence with 3n number of elements the sum of the last 2n elements is twice as big as the sum of the first 2n elements. And the sum of the first 6 elements is 252. Find the first element in this sequence. I actually encountered this problem in my math test I wasn't able to solve it so after the test I went home and tried to solve it to no avail. Any help would be appreciated!

Answer 1

If $(a+kd),\;k=1,2,\ldots,3n$

is the arithmetic progression then

the sum of the first $2n$ elements is $n(2 a + d + 2 d n)$

the sum of the last $2n$ elements is $n (2 a + d + 4 d n)$

and is given that this last sum is twice the previous one so

$n (2 a + d + 4 d n)=2n(2 a + d + 2 d n)$

simplify $n$

$2 a + d + 4 d n=2(2 a + d + 2 d n)\quad(*)$

Furthermore we know that the first $6$ element sum is $252$

$6 a + 21 d = 252$

simplify by $3$

$2a+ 7d = 84\to 2a=84-7d$

substitute in $(*)$

$84-7d + d + 4 d n=2(84-7d + d + 2 d n)$

$6 (-14 + d)=0\to d=14$

$a=84-7\cdot 14=-7$

so the progression is

$a_k=-7+14k,\;k=1,2,\ldots,3n$

and the first element is $\color{red}{a_1=7}$

Hope this is useful