Question

A TV manufacturer has produced $1000$ TVs in the seventh year and $1450$ TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the ${15}^{th}$ year

Answer 1

It is given that a TV manufacturer has produced $1000$ TV's in the seventh year that is the $7$th term of an A.P is $T_7=1000$ and also he produced $1450$ TV's in the tenth year that is the $10$th term is $T_{10}=1450$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore,

$T_7=a+(7-1)d\Rightarrow 1000=a+6d.....\left(1\right)$

$T_{10}=a+(10-1)d\Rightarrow 1450=a+9d.....\left(2\right)$

Now subtract equation 1 from equation 2 as follows:

$(a-a)+(9d-6d)=1450-1000$

$\Rightarrow 3d=450$

$\Rightarrow d=\frac{450}{3}$

$\Rightarrow d=150$

Substitute the value of $d$ in equation 1:

$a+(6\times 150)=1000$

$\Rightarrow a+900=1000$

$\Rightarrow a=1000-900$

$\Rightarrow a=100$

Therefore, the number of TV's produced in the first year is $100$.

Now, to find the number of TV's produced in the $15$th year, we have to find the $15$th term of the A.P with $a=100$ and $d=150$ is:

$T_{15}=100+(15-1)150=100+(14\times 150)=100+2100=2200$

Hence, the number of TV's produced in the $15$th year is $2200$.