Question

how do you find both the 1st term of an arithmetic sequence and the common difference?
for example, when the seventh Term in the sequence is 5 and the sum of the first 16 terms is 20​

Answer 1

An arithmetic sequence is one in which consecutive terms differ by some fixed number. Recursively, such a sequence is given by

$\begin{cases}a_1=a_1\\a_n=a_{n-1}+d&\text{for }n>1\end{cases}$

where $d$ is the common difference between terms.

Using this definition, you can express any term $a_n$ of the sequence in terms of $a_1$, since

$a_n=a_{n-1}+d$

$a_n=(a_{n-2}+d)+d=a_{n-2}+2d$

$a_n=(a_{n-3}+d)+2d=a_{n-3}+3d$

and so on down to the explicit rule,

$a_n=a_1+(n-1)d$

The sum of the first $k$ terms, $S_k$, of such a sequence is

$\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+\cdots+a_{k-1}+a_k$

or, using the explicit rule,

$\displaystyle S_k=\sum_{n=1}^k(a_1+(n-1)d)=\sum_{n=1}^k(a_1-d)+d\sum_{n=1}^kn$

$\implies\displaystyle S_k=k(a_1-d)+\frac{dk(k+1)}2=\frac{k(2a_1+(k-1)d)}2$

Now, given $a_7=5$ and $S_{16}=20$, we have

$a_7=a_1+6d\implies a_1+6d=5$

$S_{16}=\dfrac{16(2a_1+15d)}2\implies 16a_1+120d=20$

which you can solve to get

$\boxed{a_1=20,d=-\dfrac52}$