Question

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) 6 years.

[Hint: a + 5d = 8000 and a + 8d = 11300.
           Production in 8th year = (a + 7d)
           Production in 6 years = `6/2 xx [2a + 5d]`.]

Answer 1

Let production in the first year be a and the uniform yearly increase be d.

Given $$a + 5d = 8000$$ and $$a + 8d = 11300$$. 

Subtracting: $$3d = 3300 \Rightarrow d = 1100$$. 

Then $$a = 2500$$. 

(i) First year production = 2500 sets. 

(ii) 8th year = $$a + 7d$$

$$= 2500 + 7700$$

$$= 10200$$ sets

(iii) Total production in first 6 years = $$S_6$$

$$= \frac{6}{2}[2a + 5d]$$ 

$$= 3(5000 + 5500)$$

$$= 31500$$ sets