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

Solution:

Given:

Let the first term of the A.P. be a and the common difference be d.

$a_{4}=3 a$

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

$\Rightarrow a+3 d=3 a$

$\Rightarrow 3 d=2 a$

$\Rightarrow a=\frac{3 d}{2} \quad \ldots(\mathrm{i})$

And, $a_{7}-2 a_{3}=1$$\Rightarrow a+(7-1) d-2[a+(3-1) d]=1$

$\Rightarrow \mathrm{a}+6 d-2(a+2 d)=1$

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

$\Rightarrow-a+2 d=1$

$\Rightarrow-\frac{3 d}{2}+2 d=1 \quad[$ From (i) $]$

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

$\Rightarrow \frac{d}{2}=1$

$\Rightarrow d=2$

Putting the value in (i), we get:

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

$\Rightarrow a=3$