A manufacturer of TV sets produced 600 sets i | vedclass

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Question

(D) Since the production increases uniformly by a fixed number every year,the number of $TV$ sets manufactured in the $1^{st}, 2^{nd}, 3^{rd}, \dots$

Vedclass · Accepted Answer

$550$

Vedclass · Answer

(D) Since the production increases uniformly by a fixed number every year,the number of $TV$ sets manufactured in the $1^{st}, 2^{nd}, 3^{rd}, \dots$ years forms an Arithmetic Progression $(AP)$.Let $a$ be the production in the $1^{st}$ year and $d$ be the uniform increase in production every year.The production in the $n^{th}$ year is given by $a_n = a + (n-1)d$.Given,$a_3 = 600$ and $a_7 = 700$.Thus,$a + 2d = 600$ (Equation $1$) and $a + 6d = 700$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(a + 6d) - (a + 2d) = 700 - 600$.$4d = 100$,which gives $d = 25$.Substituting $d = 25$ in Equation $1$: $a + 2(25) = 600$.$a + 50 = 600$,so $a = 550$.Therefore,the production in the $1^{st}$ year is $550$ sets.

$A$ 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 production in the $1^{st}$ year.

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