# A manufacturer of TV sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find $(i)$ the production in 10th year $(ii)$ the total production in the first 7 years.

Last updated date: 21st Mar 2023

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Answer

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Hint: The production of TV is increasing by a fixed same amount every year, the production of TV in subsequent years will form an increasing A.P.

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 A.P. $a$ denotes the production in first year and $d$ denotes the number of units by which the production increases every year. So, we have:

$ \Rightarrow {a_3} = 600{\text{ and }}{a_7} = 700$.

We know that the general term of A.P. is ${T_n} = a + \left( {n - 1} \right)d$. Applying this for above equation, we have:

$

\Rightarrow a + \left( {3 - 1} \right)d = 600, \\

\Rightarrow a + 2d = 600 .....(i) \\

\Rightarrow a + \left( {7 - 1} \right)d = 700, \\

\Rightarrow a + 6d = 700 .....(ii) \\

$

Subtracting equation $(i)$ from equation $(ii)$, we’ll get:

$

\Rightarrow a + 6d - a - 2d = 700 - 600, \\

\Rightarrow 4d = 100, \\

\Rightarrow d = 25 \\

$

Putting the value of $d$ in equation $(i)$, we’ll get:

$

\Rightarrow a + 50 = 600, \\

\Rightarrow a = 550 \\

$

So, the first year’s production is 550 units and it's increasing by 25 units every year.

$(i)$ The production in 10th year will be ${a_{10}}$ which is:

$

\Rightarrow {a_{10}} = a + 9d, \\

\Rightarrow {a_{10}} = 550 + 9 \times 25, \\

\Rightarrow {a_{10}} = 550 + 225, \\

\Rightarrow {a_{10}} = 775 \\

$

So, 10th years production is 775 units.

$(ii)$ Total production in the first 7 years will be the sum of the first 7 terms of A.P.

We know that the sum of first $n$ terms of A.P. is given as:

$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right].$

Here we have, $a = 550,d = 25{\text{ and }}n = 7$. Putting these values, we’ll get:

$

\Rightarrow {S_7} = \dfrac{7}{2}\left[ {2 \times 550 + \left( {7 - 1} \right) \times 25} \right], \\

\Rightarrow {S_7} = \dfrac{{7 \times \left( {1100 + 150} \right)}}{2}, \\

\Rightarrow {S_7} = 7 \times 625, \\

\Rightarrow {S_7} = 4375 \\

$

Thus the total production in the first 7 years is 4375 units.

Note:

Sum of first 7 terms of an A.P. can also be calculated using:

\[ \Rightarrow {S_7} = \left( {\dfrac{{{a_1} + {a_7}}}{2}} \right) \times 7\] which can be conceived as

${S_7} = \left( {\dfrac{{{\text{First term + Seventh term}}}}{2}} \right) \times {\text{Number of terms}}$

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 A.P. $a$ denotes the production in first year and $d$ denotes the number of units by which the production increases every year. So, we have:

$ \Rightarrow {a_3} = 600{\text{ and }}{a_7} = 700$.

We know that the general term of A.P. is ${T_n} = a + \left( {n - 1} \right)d$. Applying this for above equation, we have:

$

\Rightarrow a + \left( {3 - 1} \right)d = 600, \\

\Rightarrow a + 2d = 600 .....(i) \\

\Rightarrow a + \left( {7 - 1} \right)d = 700, \\

\Rightarrow a + 6d = 700 .....(ii) \\

$

Subtracting equation $(i)$ from equation $(ii)$, we’ll get:

$

\Rightarrow a + 6d - a - 2d = 700 - 600, \\

\Rightarrow 4d = 100, \\

\Rightarrow d = 25 \\

$

Putting the value of $d$ in equation $(i)$, we’ll get:

$

\Rightarrow a + 50 = 600, \\

\Rightarrow a = 550 \\

$

So, the first year’s production is 550 units and it's increasing by 25 units every year.

$(i)$ The production in 10th year will be ${a_{10}}$ which is:

$

\Rightarrow {a_{10}} = a + 9d, \\

\Rightarrow {a_{10}} = 550 + 9 \times 25, \\

\Rightarrow {a_{10}} = 550 + 225, \\

\Rightarrow {a_{10}} = 775 \\

$

So, 10th years production is 775 units.

$(ii)$ Total production in the first 7 years will be the sum of the first 7 terms of A.P.

We know that the sum of first $n$ terms of A.P. is given as:

$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right].$

Here we have, $a = 550,d = 25{\text{ and }}n = 7$. Putting these values, we’ll get:

$

\Rightarrow {S_7} = \dfrac{7}{2}\left[ {2 \times 550 + \left( {7 - 1} \right) \times 25} \right], \\

\Rightarrow {S_7} = \dfrac{{7 \times \left( {1100 + 150} \right)}}{2}, \\

\Rightarrow {S_7} = 7 \times 625, \\

\Rightarrow {S_7} = 4375 \\

$

Thus the total production in the first 7 years is 4375 units.

Note:

Sum of first 7 terms of an A.P. can also be calculated using:

\[ \Rightarrow {S_7} = \left( {\dfrac{{{a_1} + {a_7}}}{2}} \right) \times 7\] which can be conceived as

${S_7} = \left( {\dfrac{{{\text{First term + Seventh term}}}}{2}} \right) \times {\text{Number of terms}}$

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