Read the text carefully and answer the questions:
Elpis Technology is a TV manufacturer company. It produces smart TV sets not only for the Indian market but also exports them to many foreign countries. Their TV sets have been in demand every time but due to the Covid-19 pandemic, they are not getting sufficient spare parts, especially chips to accelerate the production. They have to work in a limited capacity due to the lack of raw materials.

(i) They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find an increase in the production of TV every year.
(ii) They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find in which year production of TV is 1000.
OR
They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the total production in first 7 years.
(iii) They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production in the 10 th year.

(i) Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and $d$ be the common difference of the A.P. formed i.e., 'a'
denotes the production in the first year and $d$ denotes the number of units by which the production increases every year. We have, $a_3=600$ and
$
\begin{aligned}
& \mathrm{a}_3=600 \\
& \Rightarrow 600=\mathrm{a}+2 \mathrm{~d} \\
& \Rightarrow \mathrm{a}=600-2 \mathrm{~d} \ldots \\
& \Rightarrow \mathrm{a}_7=700 \\
& \Rightarrow \mathrm{a}_7=700 \\
& \Rightarrow 700=\mathrm{a}+6 \mathrm{~d} \\
& \Rightarrow \mathrm{a}=700-6 \mathrm{~d} \ldots
\end{aligned}
$
From (i) and (ii)
$
\begin{aligned}
& 600-2 d=700-6 \mathrm{~d} \\
& \Rightarrow 4 d=100 \\
& \Rightarrow d=25
\end{aligned}
$
(ii) Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and $d$ be the common difference of the A.P. formed i.e., ' $a$ ' denotes the production in the first year and $d$ denotes the number of units by which the production increases every year. We know that first term $=\mathrm{a}=550$ and common difference $=\mathrm{d}=25$
$
\begin{aligned}
& a_n=1000 \\
& \Rightarrow 1000=a+(n-1) d \\
& \Rightarrow 1000=550+25 n-25 \\
& \Rightarrow 1000-550+25=25 n \\
& \Rightarrow 475=25 n \\
& \Rightarrow n=\frac{475}{25}=19
\end{aligned}
$
OR
Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and $d$ be the common difference of the A.P. formed i.e., ' $a$ ' denotes the production in the first year and $d$ denotes the number of units by which the production increases every year. Total production in 7 years $=$ Sum of 7 terms of the A.P. with first term a $(=550)$ and $\mathrm{d}(=25)$.
$
\begin{aligned}
& S_n=\frac{n}{2}[2 a+(n-1) d] \\
& \Rightarrow S_7=\frac{7}{2}[2 \times 550+(7-1) 25] \\
& \Rightarrow S_7=\frac{7}{2}[2 \times 550+(6) \times 25] \\
& \Rightarrow S_7=\frac{7}{2}[1100+150]
\end{aligned}
$

$
\Rightarrow S_7=4375
$
(iii)Since the production increases uniformly by a fixed number every year. Therefore, the sequence formed by the production in different years is an A.P. Let a be the first term and $d$ be the common difference of the A.P. formed i.e., 'a' denotes the production in the first year and denotes the number of units by which the production increases every year. The production in the 10 th term is given by $a_{10}$. Therefore, production in the 10 th year $=a_{10}=a+9 d=550+9 \times 25=$ 775. So, production in 10 th year is of 775 TV sets.