Question

How many seven digit numbers are there in total?

Answer 1

In the given problem, we are required to compute the total number of seven digit numbers. The given question involves the concepts of arithmetic progression. The first term and last term of the arithmetic progression can be found easily. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series. For finding out the number of terms that correspond to a given term in an arithmetic progression, we must know the formula for a general term in an AP: ${a_n} = a + \left( {n - 1} \right)d$.

Complete step by step answer:
So, we have to find the number of seven digit numbers in total.
So, we have the least seven digit number as $10,00,000$.
Largest number seven digit number is $99,99,999$.
So, we have the series as: $1000000$, $1000001$, $1000002$, …… $9999999$.
We can notice that the difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression. So, we will calculate the number of terms in the AP to find the number of seven digit numbers in total.

Now, we have to find the number of terms in the AP that corresponds to $9999999$.
Here, first term $= a = 10,00,000$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio $= d = 1000001 - 1000000 = 1$
So, $d = 1$ .
We know the formula for the general term in an arithmetic progression is,
${a_n} = a + \left( {n - 1} \right)d$.
So, considering $9999999$ as the nth term of the AP, we can find the number of terms in the AP that corresponds to $124$.
So, ${a_n} = a + \left( {n - 1} \right)d = 9999999$

Substituting the values of $a$ and $d$ in the above equation, we get,
$\Rightarrow 1000000 + \left( {n - 1} \right)1 = 9999999$
Now, in order to solve the above equation for the value of n, we shift all the constant terms to the right side of the equation. So, we get,
$\Rightarrow \left( {n - 1} \right)1 = 9999999 - 1000000$
Doing the calculations, we get,
$\Rightarrow \left( {n - 1} \right) = 8999999$
$\therefore n = 8999999 + 1 = 90,00,000$

So, there are a total of $90,00,000$ seven digit numbers.

Note: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. Any term of an arithmetic progression can be calculated if we know the first term and the common difference of the arithmetic series as: ${a_n} = a + \left( {n - 1} \right)d$.