
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.
Answer
631.5k+ views
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
${S_7} = \left( {\dfrac{{{\text{First term + Seventh term}}}}{2}} \right) \times {\text{Number of terms}}$
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

The largest wind power cluster is located in the state class 11 social science CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

Which among the following are examples of coming together class 11 social science CBSE

