Solution:

Given that,

In the third and seventh year 600 and 700 radio sets units are produced, respectively.

$a_3 = 600$ and $a_7 = 700$

(і) The production in the first year

Find the production in the first year.

Let ‘$a$’ be first year production

The AP formed so is, $a$, $a+x$, $a+2x, \dots \dots$.

Using the formula,
$a_n = a + (n-1)d$

$a_3 = a + (3-1)d$

$600 = a + 2d \dots \dots (i)$

$a_7 = a + (7-1)d$

$700 = a + 6d$

$a = 700 – 6d \dots \dots (ii)$

On substituting value of a in eq.(i) we obtain,

$600 = a + 2d$

$600 = 700 – 6d + 2d$

$700 – 600 = 4d$

$100 = 4d$

$d =\frac {100}{4}$

$= 25$

Substituting the value of $d$ in eq.(ii) we obtain,

$a = 700 – 6d$

$= 700 – 6(25)$

$= 700 – 150$

$= 550$

As a result, the production in the first year, ‘a’ is 550

(ii) The total product in the 7 years
Find the total product in 7 years i.e. is $S_{7}$
Using the formula,
$\begin{array}{l}
S_{n}=n / 2[2 a+(n-1) d] \\
n=7, a=550, d=25 \\
S_{7}=7 / 2[2(550)+(7-1) 25] \\
=7 / 2[1100+150] \\
=7 / 2[1250] \\
=7[625] \\
=4375
\end{array}$
As a result, the total product in the 7 years is 4375 .