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Question: A manufacturer of TV sets produced 600 units in the third year and 700 units in the seventh year. As...

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)(i) the production in 10th year (ii)(ii) the total production in the first 7 years.

Explanation

Solution

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 aa be the first term and dd be the common difference of A.P. aa denotes the production in first year and dd denotes the number of units by which the production increases every year. So, we have:
a3=600 and a7=700\Rightarrow {a_3} = 600{\text{ and }}{a_7} = 700.
We know that the general term of A.P. is Tn=a+(n1)d{T_n} = a + \left( {n - 1} \right)d. Applying this for above equation, we have:
a+(31)d=600, a+2d=600.....(i) a+(71)d=700, a+6d=700.....(ii)  \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)(i) from equation (ii)(ii), we’ll get:
a+6da2d=700600, 4d=100, d=25  \Rightarrow a + 6d - a - 2d = 700 - 600, \\\ \Rightarrow 4d = 100, \\\ \Rightarrow d = 25 \\\
Putting the value of dd in equation (i)(i), we’ll get:
a+50=600, a=550  \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)(i) The production in 10th year will be a10{a_{10}} which is:
a10=a+9d, a10=550+9×25, a10=550+225, a10=775  \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)(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 nn terms of A.P. is given as:
Sn=n2[2a+(n1)d].\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right].
Here we have, a=550,d=25 and n=7a = 550,d = 25{\text{ and }}n = 7. Putting these values, we’ll get:
S7=72[2×550+(71)×25], S7=7×(1100+150)2, S7=7×625, S7=4375  \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:
S7=(a1+a72)×7\Rightarrow {S_7} = \left( {\dfrac{{{a_1} + {a_7}}}{2}} \right) \times 7 which can be conceived as
S7=(First term + Seventh term2)×Number of terms{S_7} = \left( {\dfrac{{{\text{First term + Seventh term}}}}{2}} \right) \times {\text{Number of terms}}