The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1.

In the given problem, let us take the first term as a and the common difference as d

Here, we are given that,

$a_{4}=3 a$.......(1)

$a_{7}=2 a_{3}+1$........(2)

We need to find a and d

So, as we know,

$a_{s}=a+(n-1) d$

For the $4^{\text {th }}$ term $(n=4)$,

$a_{4}=a+(4-1) d$

$3 a=a+3 d$

$3 a-a=3 d$

$2 a=3 d$

$a=\frac{3}{2} d$

Similarly, for the 3rd term (n = 3),

$a_{3}=a+(3-1) d$

$=a+2 d$

Also, for the 7th term (n = 7),

$a_{7}=a+(7-1) d$

$=a+6 d$ .......(3)

Now, using the value of a3 in equation (2), we get,

$a_{7}=2(a+2 d)+1$

$=2 a+4 d+1$ ........(4)

Equating (3) and (4), we get,

$a+6 d=2 a+4 d+1$

$6 d-4 d-2 a+a=+1$

$2 d-a=+1$

$2 d-\frac{3}{2} d=1$ $\left(a=\frac{3}{2} d\right)$

On further simplification, we get,

$\frac{4 d-3 d}{2}=1$

$d=(1)(2)$

$d=2$

Now, to find a,

$a=\frac{3}{2} d$

$a=\frac{3}{2}(2)$

$a=3$

Therefore, for the given A.P $d=2, a=3$