A manufacturer of TV sets produced 6000 units in the third year and 7000

Solution:

Hint: - In the question it is mentioned that the production increases by a fixed number every year.

So it is an A.P. $\left(a_{1}, a_{2}, a_{3}, a_{4}, \ldots \ldots \ldots a_{n-1}, a_{n}\right)$.

Given: -

The $3^{\text {rd }}$ year production is 6000 units

So

$a_{3}=6000^{a_{3}}=6000$

We know that $a_{n}=a+(n-1) \times d$

$a_{3}=a+(3-1) \times d$

The 7th year production is 7000 units

So,

$a_{7}=7000$

$a_{7}=a+(7-1) \times d$

$7000=a+6 d \longrightarrow(2)$

From equations (1)&(2) we get,

6000 - 2d = 7000 - 6d

4×d = 1000

$d=250 \rightarrow(3)$

From equations (1)&(2) we get,

a = 5500

i. Production in the first year $=a=5500$

$\therefore 5500$ units were produced by the manufacturer of TV sets in the first year.

ii. Production in the $10^{\text {th }}$ year $=a_{10}=a+(10-1) \times d$

$a_{10}=5500+(9) \times 250$

= 7750

$\therefore 7750$ units were produced by the manufacturer of TV sets in the $10^{\text {th }}$ year.

iii. Total production in seven years $=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}$

$s_{7}=\frac{7}{2}[2 \times a+(n-1) \times d]$

$s_{7}=\frac{7}{2}[2 \times 5500+(6) \times 250]$

$s_{7}=43750$

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