Given that the first term of an arithmetic progression is $5$ and the sum of the first eight terms is equal to the sum of the following four terms, find the common difference.
What do they mean by " sum of the first eight terms is equal to the sum of the following four terms " ?
Can you show the equation??
Is it something like $S_8=S_{12}-S_4$ ? I am stuck here
Sorry my english is terrible
On
$a=5$ sum of the first 8 terms: $S_8=\frac{n}{2}(2a+(n-1)d)$ where $n=8$
sum of the following four terms: $T_9+T_{10}+T_{11}+T_{12}$
By sum of the first eight terms is equal to the sum of the following four terms, we have:
$\frac{n}{2}(2a+(n-1)d)=T_9+T_{10}+T_{11}+T_{12}$
$\frac{8}{2}(10+(8-1)d)=a+8d+a+9d+a+10d+a+11d$
$(40+28d)=(20+38d)$
solving for d:
$20=10d$ therefore $d=2$
It means $\sum\limits_{i = 1}^8 a_i = \sum\limits_{i = 9}^{12}a_i$, where $a_i$ is $i$th term given by $a_i = a_1 + (i-1)d$, and $a_1 = 5$.