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Hint: The given question is related to arithmetic progression. Try to recall the formulae related to ${{n}^{th}}$ term of an arithmetic progression.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Letâ€™s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, in the question, we are given two arithmetic progressions with same common difference.

Let the first term of one of the progressions is ${{a}_{1}}$ and the first term of second progression is ${{b}_{1}}$ and their common difference is $d$.

Now, we will consider the first arithmetic progression.

The ${{n}^{th}}$ term of the first arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$.

So, the ${{100}^{th}}$ term of first arithmetic progression is given as ${{a}_{100}}={{a}_{1}}+99d........(i)$.

Now, we will consider the second arithmetic progression.

The ${{n}^{th}}$ term of the second arithmetic progression is given as ${{b}_{n}}={{b}_{1}}+(n-1)d$.

So, the ${{100}^{th}}$ term of second arithmetic progression is given as ${{b}_{100}}={{b}_{1}}+99d........(ii)$.

Now, in the question, we are given that the difference between the ${{100}^{th}}$ term of both the arithmetic progressions is equal to $111222333$.

So, from $(i)$ and $(ii)$, we can say $\left( {{a}_{1}}+99d \right)-\left( {{b}_{1}}+99d \right)=111222333$

$\Rightarrow {{a}_{1}}-{{b}_{1}}=111222333.....(iii)$

Now, we are asked to find the difference between their millionth term. The millionth term of the first arithmetic progression is given as ${{a}_{1000000}}={{a}_{1}}+999999d$ and the millionth term of the second arithmetic progression is given as ${{b}_{1000000}}={{b}_{1}}+999999d$ .

We need to find their difference. The difference is given as ${{a}_{1000000}}-{{b}_{1000000}}$.

$=\left( {{a}_{1}}+999999d \right)-\left( {{b}_{1}}+999999d \right)$

$={{a}_{1}}-{{b}_{1}}$

Now , from equation$(iii)$ , we have ${{a}_{1}}-{{b}_{1}}=111222333$.

So, ${{a}_{1000000}}-{{b}_{1000000}}=111222333$.

Hence, the difference between the millionth term of the given two arithmetic progressions is equal to $111222333$.

Note: Generally, students get confused in the expression for ${{n}^{th}}$ term of an arithmetic progression. The ${{n}^{th}}$ term of the arithmetic progression with first term ${{a}_{1}}$ and common difference $d$ is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ and not ${{a}_{n}}={{a}_{1}}+nd$ . This confusion should be avoided as they can result in getting wrong answers.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Letâ€™s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, in the question, we are given two arithmetic progressions with same common difference.

Let the first term of one of the progressions is ${{a}_{1}}$ and the first term of second progression is ${{b}_{1}}$ and their common difference is $d$.

Now, we will consider the first arithmetic progression.

The ${{n}^{th}}$ term of the first arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$.

So, the ${{100}^{th}}$ term of first arithmetic progression is given as ${{a}_{100}}={{a}_{1}}+99d........(i)$.

Now, we will consider the second arithmetic progression.

The ${{n}^{th}}$ term of the second arithmetic progression is given as ${{b}_{n}}={{b}_{1}}+(n-1)d$.

So, the ${{100}^{th}}$ term of second arithmetic progression is given as ${{b}_{100}}={{b}_{1}}+99d........(ii)$.

Now, in the question, we are given that the difference between the ${{100}^{th}}$ term of both the arithmetic progressions is equal to $111222333$.

So, from $(i)$ and $(ii)$, we can say $\left( {{a}_{1}}+99d \right)-\left( {{b}_{1}}+99d \right)=111222333$

$\Rightarrow {{a}_{1}}-{{b}_{1}}=111222333.....(iii)$

Now, we are asked to find the difference between their millionth term. The millionth term of the first arithmetic progression is given as ${{a}_{1000000}}={{a}_{1}}+999999d$ and the millionth term of the second arithmetic progression is given as ${{b}_{1000000}}={{b}_{1}}+999999d$ .

We need to find their difference. The difference is given as ${{a}_{1000000}}-{{b}_{1000000}}$.

$=\left( {{a}_{1}}+999999d \right)-\left( {{b}_{1}}+999999d \right)$

$={{a}_{1}}-{{b}_{1}}$

Now , from equation$(iii)$ , we have ${{a}_{1}}-{{b}_{1}}=111222333$.

So, ${{a}_{1000000}}-{{b}_{1000000}}=111222333$.

Hence, the difference between the millionth term of the given two arithmetic progressions is equal to $111222333$.

Note: Generally, students get confused in the expression for ${{n}^{th}}$ term of an arithmetic progression. The ${{n}^{th}}$ term of the arithmetic progression with first term ${{a}_{1}}$ and common difference $d$ is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ and not ${{a}_{n}}={{a}_{1}}+nd$ . This confusion should be avoided as they can result in getting wrong answers.

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