The production of a $TV$ factory increases equally every year. Its production in the $3^{rd}$ year was $600 \, TV$ and its production in the $7^{th}$ year was $700 \, TV$. Find the production in the $1^{st}$ year,in the $10^{th}$ year,and the total production in $7$ years.

(N/A) Since the production of the factory increases equally every year,the annual production forms an Arithmetic Progression $(A.P.)$.
For this $A.P.$,the $3^{rd}$ term $T_{3} = 600$ and the $7^{th}$ term $T_{7} = 700$.
The common difference $d$ is given by $d = \frac{T_{m} - T_{n}}{m - n}$.
Substituting $m = 7$ and $n = 3$,we get:
$d = \frac{700 - 600}{7 - 3} = \frac{100}{4} = 25$.
Using the formula $T_{n} = a + (n - 1)d$ for $T_{3}$:
$600 = a + (3 - 1)(25) \implies 600 = a + 50 \implies a = 550$.
Thus,the production in the $1^{st}$ year is $550 \, TV$.
For the $10^{th}$ year:
$T_{10} = a + 9d = 550 + 9(25) = 550 + 225 = 775 \, TV$.
For the total production in $7$ years $(S_{7})$:
$S_{n} = \frac{n}{2} [2a + (n - 1)d]$
$S_{7} = \frac{7}{2} [2(550) + (7 - 1)(25)] = \frac{7}{2} [1100 + 150] = \frac{7}{2} [1250] = 7 \times 625 = 4375 \, TV$.