
A manufacturer of TV sets produced 720 sets in fourth year and 1080 sets in the sixth year. Assuming that the production increases uniformly by a fixed number every year,then finds the total production in the first 9 years.
Answer
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Hint: Here, we will be using the formulas for last term of an AP and sum of first terms of an AP series.AP means arithmetic progression.Since the production increases uniformly by a fixed number every year,then this will form an AP.
Complete step-by-step answer:
Let us assume that production in first year is $a$ and each year $d$ number of sets are increased.This will form an AP like $a,a + d,a + 2d,a + 3d,.....$
Step1:Hence, for the n-th year the production will be ${t_n} = a + (n - 1)d$ ,n-th term of an AP series.
Step2:The total production for the n years will be ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$ ,sum of first n terms of an AP series.
Step3:Here,fourth year production $or,{t_4} = 720$ and sixth year production ${t_6} = 1080$ Step4:Then we get two equations:
${t_4} = 720$
$a + (4 - 1)d = 720$
$or,\,\,a + 3d = 720$ ………………(1)
Step5:Again,
${t_6} = 1080$
$or,\,\,a + (6 - 1)d = 1080$
$or,\,\,a + 5d = 1080$ .......................(2)
Now we have to solve equations (1) and (2).
Step6:Subtracting (1) from (2),
$a + 5d - (a + 3d) = 1080 - 720$
$\eqalign{
& or,\,\,5d - 3d = 360 \cr
& or,\,\,2d = 360 \cr
& or,\,\,d = 180 \cr} $
Step7:So,we find the value of d.Putting the value of d in equation (1),
$\eqalign{
& a + 3 \times 180 = 720 \cr
& or,\,\,a = 720 - 540 \cr
& or,\,\,a = 180 \cr} $
Step8:Now, the total production in the first 9 years that is ,
$\eqalign{
& {S_9} = \dfrac{9}{2}[2 \times a + (9 - 1)d] \cr
& \,\,\,\,\,\, = \dfrac{9}{2}[2 \times 180 + 8 \times 180] \cr
& \,\,\,\,\,\, = \dfrac{9}{2} \times 180 \times 10 \cr
& \,\,\,\,\,\, = 45 \times 180 \cr
& \,\,\,\,\,\,\, = 8100 \cr} $
Therefore,the total production for the first 9 years will be 8100 sets.
Note: In the Problem, the number of productions increases by a fixed number uniformly. For that reason,we take it as a constant. Remember the sum of n term formulas of AP for easy calculation.
Complete step-by-step answer:
Let us assume that production in first year is $a$ and each year $d$ number of sets are increased.This will form an AP like $a,a + d,a + 2d,a + 3d,.....$
Step1:Hence, for the n-th year the production will be ${t_n} = a + (n - 1)d$ ,n-th term of an AP series.
Step2:The total production for the n years will be ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$ ,sum of first n terms of an AP series.
Step3:Here,fourth year production $or,{t_4} = 720$ and sixth year production ${t_6} = 1080$ Step4:Then we get two equations:
${t_4} = 720$
$a + (4 - 1)d = 720$
$or,\,\,a + 3d = 720$ ………………(1)
Step5:Again,
${t_6} = 1080$
$or,\,\,a + (6 - 1)d = 1080$
$or,\,\,a + 5d = 1080$ .......................(2)
Now we have to solve equations (1) and (2).
Step6:Subtracting (1) from (2),
$a + 5d - (a + 3d) = 1080 - 720$
$\eqalign{
& or,\,\,5d - 3d = 360 \cr
& or,\,\,2d = 360 \cr
& or,\,\,d = 180 \cr} $
Step7:So,we find the value of d.Putting the value of d in equation (1),
$\eqalign{
& a + 3 \times 180 = 720 \cr
& or,\,\,a = 720 - 540 \cr
& or,\,\,a = 180 \cr} $
Step8:Now, the total production in the first 9 years that is ,
$\eqalign{
& {S_9} = \dfrac{9}{2}[2 \times a + (9 - 1)d] \cr
& \,\,\,\,\,\, = \dfrac{9}{2}[2 \times 180 + 8 \times 180] \cr
& \,\,\,\,\,\, = \dfrac{9}{2} \times 180 \times 10 \cr
& \,\,\,\,\,\, = 45 \times 180 \cr
& \,\,\,\,\,\,\, = 8100 \cr} $
Therefore,the total production for the first 9 years will be 8100 sets.
Note: In the Problem, the number of productions increases by a fixed number uniformly. For that reason,we take it as a constant. Remember the sum of n term formulas of AP for easy calculation.
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