The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 8000 sets in 6th year, and 11300 in 9th year. Find the production in:

(i) first year

(ii) 8th year

(iii) total production in 6 years.

(i) Given,

The production of TV sets in a factory increases uniformly by a fixed number every year. So the production is in A.P.

Let the TV sets produced in first year be a and difference in production of each year be d.

We know that,

∴ a_n = a + (n - 1)d

Given,

The sets produced in 6 years is 8000.

⇒ a + (6 - 1)d = 8000

⇒ a + 5d = 8000 ….(1)

The sets produced in the 9th year is 11300.

⇒ a + (9 - 1)d = 11300

⇒ a + 8d = 11300 ….(2)

Subtract Equation (1) from Equation (2), we get:

⇒ a + 8d - (a + 5d) = 11300 - 8000

⇒ 3d = 3300

⇒ d = $\dfrac{3300}{3}$

⇒ d = 1100.

Substitute d = 1100 into Equation 1 :

⇒ a + 5(1100) = 8000

⇒ a + 5500 = 8000

⇒ a + 5500 = 8000

⇒ a = 8000 - 5500

⇒ a = 2500.

Hence, the production in the first year is 2500 sets.

(ii) Solving,

⇒ a₈ = a + 7d

= 2500 + 7(1100)

= 2500 + 7700

= 10200.

Hence, production in 8th year = 10200.

(iii) Total production in 6 years :

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₆ = $\dfrac{6}{2}$ [2(2500) + (6 - 1)1100]

= 3[5000 + (5)1100]

= 3[5000 + 5500]

= 3(10500)

= 31500.

Hence, total production in 6 years = 31500.