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 1^st year.

(ii) The production in 10^th year.

(iii) The total production in 7 years. (CBSE 2015)

Let the TV sets produced in 1^st 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 3^rd year is 600 and in 7^th year is 700

⇒ 3^rd term of AP, a₃ = 600

⇒ 7^th 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

a_n = 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 1^st year is first term of AP = 550

(ii) The production in 10^th year is 10^th 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.