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Question: A manufacturer of TV sets produced 720 sets in fourth year and 1080 sets in the sixth year. Assuming...

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.

Explanation

Solution

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 aa and each year dd number of sets are increased.This will form an AP like a,a+d,a+2d,a+3d,.....a,a + d,a + 2d,a + 3d,.....
Step1:Hence, for the n-th year the production will be tn=a+(n1)d{t_n} = a + (n - 1)d ,n-th term of an AP series.
Step2:The total production for the n years will be Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}[2a + (n - 1)d] ,sum of first n terms of an AP series.
Step3:Here,fourth year production or,t4=720or,{t_4} = 720 and sixth year production t6=1080{t_6} = 1080 Step4:Then we get two equations:
t4=720{t_4} = 720
a+(41)d=720a + (4 - 1)d = 720
or,a+3d=720or,\,\,a + 3d = 720 ………………(1)

Step5:Again,
t6=1080{t_6} = 1080
or,a+(61)d=1080or,\,\,a + (6 - 1)d = 1080
or,a+5d=1080or,\,\,a + 5d = 1080 .......................(2)

Now we have to solve equations (1) and (2).
Step6:Subtracting (1) from (2),
a+5d(a+3d)=1080720a + 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.