The algebraic form of an arithmetic sequence is 5n+3
[a] what is the first term of the sequence
[b] What will be the remainder if the terms of the sequence are divided by 5.

Hint: Use the fact that if a sequence is given by $\left\{ {{a}_{n}}=f\left( n \right) \right\}$, then the first term of the sequence is determined by putting n = 1 in the formula for the general term of the sequence, i.e. $a=f\left( 1 \right)$. Hence determine the first term of the arithmetic sequence. Use the fact that if a,b are any two integers, then there exist unique integers r and q such that $a=bq+r,0\le r\le b-1$. Here r is known as the remainder of the division of a by b. Hence determine the remainder obtained on dividing ${{a}_{n}}$ by 5.

Complete step-by-step answer:
We know that if a sequence is given by $\left\{ {{a}_{n}}=f\left( n \right) \right\}$, then the first term of the sequence is determined by putting n = 1 in the formula for the general term of the sequence, i.e. $a=f\left( 1 \right)$.
Now, we have
${{a}_{n}}=5n+3$
Hence, we have
$a=5\left( 1 \right)+3=8$
Hence the first term of the arithmetic sequence is 8.
Now, we have
${{a}_{n}}=5n+3\text{ }\left( i \right)$
Also by Euclid’s division lemma, if a,b are any two integers, then there exist unique integers r and q such that $a=bq+r,0\le r\le b-1$.
Hence, we have
${{a}_{n}}=5q+r,0\le r\le 4\text{ }\left( ii \right)$
Since q and r are unique, from equations (i) and (ii), we get
n = q and r = 3.
Hence the remainder obtained on dividing the term of the sequence by 5 is 3.

Note: Alternative solution for part [a]:
We have
${{a}_{n}}=5n+3=5\left( n-1+1 \right)+3=5+3+5\left( n-1 \right)$
Hence, we have
${{a}_{n}}=8+5\left( n-1 \right)$
Comparing with the formula for the general term of an arithmetic sequence, i.e. ${{a}_{n}}=a+\left( n-1 \right)d$, we get
$a=8,d=5$
Hence, the first term of the sequence is 8.