Question

The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 8000 sets in 6th year and 11300 in 9th year. Find the production in
(i) first year
(ii) 8th year
(iii) 6 years

Answer 1

Hint: The production of TV sets increases uniformly by fixed number. So the no. of TV sets produced in every year forms an Arithmetic progression with the production in first year (first term) as ‘a’ and increases uniformly by fixed number ‘d’. So in this A.P we have 6th and 9th years production (6th and 9th terms), using this we can find the value of ‘a’ and ‘d’; so then we can find the production in 1st year, 8th year and total production in 6 years.
Formulas used:
nth term of an A.P is $ {T_n} = a + \left( {n - 1} \right)d $ and Sum of n terms of an A.P is $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ , where a is the first term, d is the common difference and n is the no. of the term.

Complete step-by-step answer:
We are given that production of TV sets in a factory increases uniformly by a fixed number every year.
It produced 8000 TV sets in 6th year and 11300 in the 9Th year.
The series of production in each year forms an A.P
In this A.P we have 6th and 9th terms.
I.For 6th term (year), n is 6, $ {T_6} = a + \left( {6 - 1} \right)d = a + 5d = 8000 $ …….. Equation (1)
For 9th term (year), n is 9, $ {T_9} = a + \left( {9 - 1} \right)d = a + 8d = 11300 $ ………. Equation (2)
On subtracting equation 1 from equation 2, we get $ a + 8d - a - 5d = 11300 - 8000 $
 $ \Rightarrow 3d = 3300 $
 $ \therefore d = \dfrac{{3300}}{3} = 1100 $
On substituting the value of d in equation 1, we get
 $ \Rightarrow a + 5\left( {1100} \right) = 8000 $
 $ \Rightarrow a = 8000 - 5500 = 2500 $
Therefore, the first term which is the production in the first year is 2500 sets.
So, the correct answer is “2500 Sets”.

(ii) Production in 8th year
For 8th year (term), n is 8
Therefore, production in 8th year is
 $ a + \left( {n - 1} \right)d = a + \left( {8 - 1} \right)d = 2500 + 7\left( {1100} \right) = 2500 + 7700 = 10,200 $ sets.
So, the correct answer is “10,200 Sets”.

(iii) Total production in 6 years.
For 6 years, n is 6
Sum of production in first 6 years is $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
 $ \Rightarrow {S_6} = \dfrac{6}{2}\left[ {2a + \left( {6 - 1} \right)d} \right] $
 $ \Rightarrow {S_6} = 3\left[ {2\left( {2500} \right) + 5\left( {1100} \right)} \right] = 3\left( {5000 + 5500} \right) $
 $ \therefore {S_6} = 3 \times 10500 = 31,500 $ sets.
So, the correct answer is “31,500 Sets”.

Note: The production increases uniformly by a fixed number. So we have considered it as an A.P. In an A.P, every term starting from second term is obtained by adding a fixed number for its previous term and this fixed number is called common difference whereas in an G.P the terms are obtained by multiplying their previous terms by a fixed ratio and this ratio is called common ratio. Always remember that, while subtracting one equation from another, subtract LHS of one equation from LHS of another equation and same with RHS and do not swap them.