Question

Let a4+a8+a12+a16=224. Find the sum of first 19 term in sequence.

Answer 1

Without knowing anything about the sequence, this is impossible to answer. But suppose the sequence is arithmetic, in which case each term differs by some constant $k$:

$a_2=a_1+k$

$a_3=a_2+k=a_1+2k$

$a_4=a_3+k=a_1+3k$

$\ldots$

$a_n=a_{n-1}+k=a_1+(n-1)k$

Then we can write

$a_4+a_8+a_{12}+a_{16}=4a_1+36k=224\implies a_1+9k=56$

and from the formula above, we see this means the 10th term in the sequence is $a_{10}=56$. But that's all the specific info we can gather about such an arithmetic sequence. If we set the first term to be some unknown $a_1=a$, then the sum of the first 19 terms in the sequence would be

$\displaystyle\sum_{n=1}^{19}a_n=\sum_{n=1}^{19}\big(a+(n-1)k\big)=19a+k\sum_{n=1}^{18}n=19a+171k$

If we knew one more term in the sequence, we could determine the value of $k$ and derive the value of $a$ (if the first term $a_1$ is not immediately given), and then go on to find an exact numeric value for the sum.