Question

The production of TV in a factory increase uniformly by a fixed number every year. It produced 8000 acts in 6th year and 11300 in 9th year. Find the production in (i) first year (ii) 8th year and (iii) 6th year.

Answer 1

(i) As the production of TV sets increases uniformly by a fixed number every year, the number of TV sets produced in the 1^st year, 2^nd year, 3^rd year, ... will form an AP. Let the first term be a, common difference of the AP be d and the number of TV sets produced in the n^th year be T_n.

Then T₆ = 8000
⇒ a + 5d = 8000 ...(i)
Also, T₉ = 11300
⇒ a + 8d = 11300 ...(ii)

On subtracting (i) from (ii), we get:
⇒ 3d = 3300
⇒​ d = 1100

Putting the value of d in (i), we get:
⇒ a + 5 ⨯ 1100 = 8000
⇒ a = 8000 - 5500 = 2500

Therefore, the production of TV sets in 1^st year is 2500.

(ii) Now, production of TV sets in the 8^th year is given by
T₈ = a + 7d = 2500 + 7 ⨯ 1100 = 10200

(iii) Sum of n terms of an AP is given by
$$\begin{gathered} S_n=\frac{n}{2}\left[a+l\right] \\ \Rightarrow S_6=\frac{6}{2}\left[2500+8000\right]=3\times 10500=31500 \end{gathered}$$
Thus, the total production of TV sets in the 6^th years is 31500.