
A manufacturer 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 the production in the
a) The first year …………… (a)
b) The 10th year……………. (b)
c) 7 years ……………………. (c)
d) Find (c)-(a)-(b)
Answer
595.8k+ views
Hint: As the production increases uniformly by a fixed number every year, the production of the units of TV in the years will be in the form of an arithmetic progression. We can use the formulas for finding the nth term of an A.P. and the sum of n terms of an A.P. to obtain the answer to the question.
Complete step-by-step answer:
It is given that production increases uniformly by a fixed number every year, let’s say by a fixed number d. Therefore, if the no. of units of TV produced in the first year is a, then the units produced in the second year will be a+d, third year will be a+2d and so on.
Therefore, the number of units produced in the years will be an arithmetic progression with initial term a and common difference d. Therefore,
The number of units of TV produced in nth year= a+(n-1)d ………………………………..(1.1)
a)It is given that the number of units of TV produced in third and seventh year is 600 and 700 respectively. Thus, form (1.1),
$a+(3-1)d=600\Rightarrow a+2d=600........(1.2)$
And $a+(7-1)d=600\Rightarrow a+6d=700.........(1.3)$
Subtracting (1.2) from (1.3), we get
$4d=100\Rightarrow d=25...............(1.4)$
Using this value of d in equation (1.2), we get
$a+2\times 25=600\Rightarrow a=550..............(1.5)$
Therefore, the number of units produced in 1st year=550
b)Using equations (1.1), (1.4) and (1.5), we get
No. of units produced in 10th year=$a+(10-1)d=550+9\times 25=775............(1.6)$
c)The sum of n terms of an AP is given by
${{s}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
Taking n=7, and using equation (1.4) and (1.5) to get a and d,
\[The\text{ }production\text{ }in\text{ }7\text{ }years=\dfrac{7}{2}\left( 2\times 550+6\times
25 \right)=4375..............(1.7)\]
d)Thus, from equations (1.5), (1.6) and (1.7), we find that
(c)-(a)-(b)= 4375-775-550 = 3050
Which is the answer to the question.
Note: We should be careful that the nth term of an A.P. is given by \[a+\left( n-1 \right)d\] and not \[a+nd\] because in the first term, d is not added and gets added only from the second term onwards.
Complete step-by-step answer:
It is given that production increases uniformly by a fixed number every year, let’s say by a fixed number d. Therefore, if the no. of units of TV produced in the first year is a, then the units produced in the second year will be a+d, third year will be a+2d and so on.
Therefore, the number of units produced in the years will be an arithmetic progression with initial term a and common difference d. Therefore,
The number of units of TV produced in nth year= a+(n-1)d ………………………………..(1.1)
a)It is given that the number of units of TV produced in third and seventh year is 600 and 700 respectively. Thus, form (1.1),
$a+(3-1)d=600\Rightarrow a+2d=600........(1.2)$
And $a+(7-1)d=600\Rightarrow a+6d=700.........(1.3)$
Subtracting (1.2) from (1.3), we get
$4d=100\Rightarrow d=25...............(1.4)$
Using this value of d in equation (1.2), we get
$a+2\times 25=600\Rightarrow a=550..............(1.5)$
Therefore, the number of units produced in 1st year=550
b)Using equations (1.1), (1.4) and (1.5), we get
No. of units produced in 10th year=$a+(10-1)d=550+9\times 25=775............(1.6)$
c)The sum of n terms of an AP is given by
${{s}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
Taking n=7, and using equation (1.4) and (1.5) to get a and d,
\[The\text{ }production\text{ }in\text{ }7\text{ }years=\dfrac{7}{2}\left( 2\times 550+6\times
25 \right)=4375..............(1.7)\]
d)Thus, from equations (1.5), (1.6) and (1.7), we find that
(c)-(a)-(b)= 4375-775-550 = 3050
Which is the answer to the question.
Note: We should be careful that the nth term of an A.P. is given by \[a+\left( n-1 \right)d\] and not \[a+nd\] because in the first term, d is not added and gets added only from the second term onwards.
Recently Updated Pages
In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

In cricket, what is a "tail-ender"?

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Who Won 36 Oscar Awards? Record Holder Revealed

The time gap between two sessions of the Parliament class 10 social science CBSE

