An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Let’s consider the first term and the common difference of the A.P to be $a$ and $d$ respectively.

Then, we know that ${{a}_{n}}=a+\left( n-1 \right)d$

Given conditions,

${{4}^{th}}$ term of an A.P. is three times the first

Expressing this by equation we have,

⇒ ${{a}_{4}}=3\left( a \right)$

$a+\left( 4-1 \right)d=3a$

$3d=2a$ ⇒ $a=3d/2$…….$\left( i \right)$

And,

${{7}^{th}}$ term exceeds twice the third term by $1$

⇒ ${{a}_{7}}=2\left( {{a}_{3}} \right)+1$

$a+\left( 7-1 \right)d=2\left( a+\left( 3-1 \right)d \right)+1$

$a-2d+1=0$ ….. $\left( ii \right)$

Using $\left( i \right)$ in $\left( ii \right),$ we have

$3d/2-2d+1=0$

$3d-4d+2=0$

$d=2$

So, putting $d=2$ in $\left( i \right),$ we get a

⇒ $a=3$

Therefore, the first term is $3$ and the common difference is $2.$