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(B) Let the production in the $n^{th}$ year be $a_n$. Since the production increases uniformly,it forms an Arithmetic Progression $(AP)$.Given: $a_3 =

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$4375$

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(B) Let the production in the $n^{th}$ year be $a_n$. Since the production increases uniformly,it forms an Arithmetic Progression $(AP)$.Given: $a_3 = 600$ and $a_7 = 700$.The formula for the $n^{th}$ term is $a_n = a + (n-1)d$.So,$a + 2d = 600$ (Equation $1$) and $a + 6d = 700$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(a + 6d) - (a + 2d) = 700 - 600$,which gives $4d = 100$,so $d = 25$.Substituting $d = 25$ in Equation $1$: $a + 2(25) = 600$,so $a + 50 = 600$,which gives $a = 550$.The total production in the first $7$ years is the sum of the first $7$ terms,$S_7$.The sum formula is $S_n = \frac{n}{2}[2a + (n-1)d]$.$S_7 = \frac{7}{2}[2(550) + (7-1)(25)] = \frac{7}{2}[1100 + 150] = \frac{7}{2}[1250] = 7 	imes 625 = 4375$.Thus,the total production of $TV$ sets in the first $7$ years is $4375$.

$A$ manufacturer of $TV$ sets 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.

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