Answer

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**Hint:**If a quantity is uniformly increasing, it should have a constant difference, d between any two consecutive values in the series. The collection of these values is known as an arithmetic progression. By identifying this difference, we can calculate an unknown value at any given instant.

**Complete Step by Step Solution:**

Let us consider an arithmetic progression with n number of terms. Let the initial term be a and the difference be d. Therefore the progression will be in the form as shown below.

$a,a+d,a+2d,....,a+\left( n-1 \right)d$ where n is an integer.

Therefore the value of the ${{p}^{th}}$ term will be equal to $a+\left( p-1 \right)d$. Using this expression let us find out the difference, d of this arithmetic progression.

It is given that in the third year, the production is 6000 units.

$\Rightarrow a+(p-1)d=6000$

For the third year, $p=3$. Therefore

$\Rightarrow a+(3-1)d=6000$

$\Rightarrow a+2d=6000$ ……(1)

For the seventh year, the production, $p=7$

$\Rightarrow a+(7-1)d=7000$

$\Rightarrow a+6d=7000$ ……(2)

Let us solve the equations (1) to (2) to a and d.

Equation (1) can be rewritten as

$\Rightarrow a=6000-2d$

Let us substitute the above data in equation (2)

$\Rightarrow 6000-2d+6d=7000$

$\Rightarrow 4d=1000\Rightarrow d=250$

Now substitute the value of the difference in (1) or (2) to get the initial value a. Let us substitute $d=250$ in (1)

$\Rightarrow a+2\times 250=6000$

$\Rightarrow a=5500$

We now have the required data to calculate ${{p}^{th}}$ term.

Let us calculate the production of TV in the tenth year $\left( p=10 \right)$.

$\Rightarrow a+\left( p-1 \right)d=5500+\left( 10-1 \right)250$

$\Rightarrow 5500+2250=7750$

**Therefore the production will be 7750 units in the tenth year.**

**Note:**

This problem can be solved intuitively without any of these formulas. If we notice the question, we can see that the production has increased by 1000 units in four years. Which is actually 250 units per year.

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