A manufacturer of TV sets produced 6000 units in the third year and 7000 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production (i) in the first year, (ii) in the 10th year, (iii) in 7 years.
A manufacturer of TV sets produced 6000 units in the third year and 7000 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production (i) in the first year, (ii) in the 10th year, (iii) in 7 years.

Hint: – In the question it is mentioned that the production increases by a fixed
number every year.
So it is an A.P. (a1, a2, a3, a4, ……..an – 1, an).
Given: –
The 3rd year production is 6000 units
So,
a3 = 6000
We know that an = a + (n – 1) ×d
a3 = a + (3 – 1)×d
6000 = a + 2d
The 7th year production is 7000 units
So,
a7 = 7000
a7 = a + (7 – 1)×d

7000 = a + 6d
From equations (1)&(2) we get,
6000 – 2d = 7000 – 6d
4×d = 1000
d = 250
From equations (1)&(2) we get,
a = 5500
i. Production in the first year = a = 5500
∴ 5500 units were produced by the manufacturer of TV sets in the first year.
ii. Production in the 10th year = a10 = a + (10 – 1)×d
a10 = 5500 + (9) ×250
= 7750
∴ 7750 units were produced by the manufacturer of TV sets in the 10th year.
iii. Total production in seven years = a1 + a2 + a3 + a4 + a5 + a6 + a7

s7=

$\frac{7}{2}\left[2a+\left(7–1\right)d\right]=\frac{7}{2}\left[2×5500+6×250\right]$

s7 = 43750
∴ A total of 16, 250 units was produced by the manufacturer in 7 years.