A manufacturer of TV sets produced 600 sets i | vedclass

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(A) Let the production in the $n^{th}$ year be represented by an arithmetic progression $a_n = a + (n-1)d$,where $a$ is the production in the first ye

Vedclass · Accepted Answer

$775$

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(A) Let the production in the $n^{th}$ year be represented by an arithmetic progression $a_n = a + (n-1)d$,where $a$ is the production in the first year and $d$ is the constant annual increase.Given: $a_3 = 600$ and $a_7 = 700$.Using the formula $a_n = a + (n-1)d$:$a + 2d = 600$ (Equation $1$)$a + 6d = 700$ (Equation $2$)Subtracting Equation $1$ from Equation $2$:$(a + 6d) - (a + 2d) = 700 - 600$$4d = 100$$d = 25$Substituting $d = 25$ into Equation $1$:$a + 2(25) = 600$$a + 50 = 600$$a = 550$Now,find the production in the $10^{th}$ year $(a_{10})$:$a_{10} = a + 9d$$a_{10} = 550 + 9(25)$$a_{10} = 550 + 225 = 775$Thus,the production in the $10^{th}$ year is $775$ 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 $10^{th}$ year.

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