Q30 of 34 Page 1

A manufacturer of TV sets produced 600 units in the 3rd year and 700 units in the 7th year. Assuming that, production increases uniformly by a fixed number of units every year. Find

(i) The production in 1st year.


(ii) The production in 10th year.


(iii) The total production in 7 years.

Let the TV sets produced in 1st year be 'a' and as the production increases uniformly by a fixed number of units every year, let that be 'd'.


Then, the TV sets produced each year are,


a, a + d, a + 2d, ….,


Clearly, the sequence of units produced each year makes an AP.


Given, that


No of TV sets produced in 3rd year is 600 and in 7th year is 700


⇒ 3rd term of AP, a3 = 600


⇒ 7th term of AP,a7 = 700


Also, if 'a' is first term and 'd' is common difference of an AP, then nth term of an AP is


an = a + (n - 1)d


As, a3 = 600


⇒ a + 2d = 600


⇒ a = 600 - 2d …[1]


Also, a7 = 700


⇒ a + 6d = 700


⇒ 600 - 2d + 6d = 700


⇒ 4d = 100


⇒ d = 25


Putting 'd' in [1]


⇒ a = 600 - 2(25) = 550


(i) The production in 1st year is first term of AP = 550


(ii) The production in 10th year is 10th term of AP.


Since,


a10 = a + 9d


a10 = 550 + 9(25) = 775


(iii) The total production in 7 years is sum of 7 terms of AP.


And we know,


Sum of an AP


Where, a is first term and d is common difference


Putting values, we get




i.e. The company produced 4375 TV sets in 7 years.


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