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

Solution:

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 $d=2, a=3$