the 4th term of an AP …

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

Now we have,
a₄ = 3a
{tex}\Rightarrow{/tex} a + (4 - 1)d = 3a
{tex}\Rightarrow{/tex}a + 3d - 3a = 0
{tex}\Rightarrow{/tex}-2a + 3d = 0............(i)
and,a₇=2a₃+1

{tex}\implies{/tex} a₇ - 2a₃ = 1
{tex}\Rightarrow{/tex} a+ (7 - 1)d -2[a + (3 - 1)d] = 1
{tex}\Rightarrow{/tex} a + 6d - 2[a + 2d] = 1 
{tex}\Rightarrow{/tex} a + 6d - 2a - 4d = 1
{tex}\Rightarrow{/tex} -a + 2d = 1 (multiplying by 2 both sides)
{tex}\Rightarrow{/tex} -2a + 4d = 2..........(ii)
Subtracting equation (i) from (ii),
(-2a + 4d) - (-2a + 3d) = 2 - 0
{tex}\Rightarrow{/tex} -2a + 4d + 2a - 3d = 2
{tex}\Rightarrow{/tex} d = 2
Put the value of d in (i)
-2a + 3{tex}\times{/tex} 2 = 0
{tex}\Rightarrow{/tex} -2a = -6
{tex}\Rightarrow \quad a = \frac { - 6 } { - 2 } = 3{/tex}
Therefore, First term a = 3 and Common difference d = 2