(A) The arithmetic mean $A$ is given by:$A = \frac{9 + 99 + 999 + \dots + 999999999}{9}$Dividing each term by $9$,we get:$A = 1 + 11 + 111 + \dots + 1

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(A) The arithmetic mean $A$ is given by:$A = \frac{9 + 99 + 999 + \dots + 999999999}{9}$Dividing each term by $9$,we get:$A = 1 + 11 + 111 + \dots + 111111111$Summing these terms:$1 = 1$$1 + 11 = 12$$1 + 11 + 111 = 123$Continuing this pattern for $9$ terms,we get:$A = 123456789$The number $N = 123456789$ consists of the digits $1$ through $9$. It does not contain the digit $0$.

Class 11 Mathematics Sequences and Series Arithmetic progression

The arithmetic mean of the nine numbers in the given set $\{9, 99, 999, \dots, 999999999\}$ is a $9$-digit number $N$,all whose digits are distinct. The number $N$ does not contain the digit:

A
$0$
B
$2$
C
$5$
D
$9$

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Sequences and Series

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Arithmetic progression

If the sum of the first $10$ terms of an arithmetic progression is $4$ times the sum of its first $5$ terms,then the ratio of its first term to the common difference is......

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If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an arithmetic sequence are $a$,$b$,and $c$ respectively,then the value of $[a(q - r) + b(r - p) + c(p - q)]$ is:

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If $a_1, a_2, ..., a_{24}$ are in an arithmetic progression and $a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225$,then what is the sum of the first $24$ terms of this arithmetic progression?

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The first term of an $A$.$P$. of $30$ non-negative terms is $\frac{10}{3}$. If the sum of this $A$.$P$. is the cube of its last term,then its common difference is:

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The arithmetic mean of the nine numbers in the given set $\{9, 99, 999, \dots, 999999999\}$ is a $9$-digit number $N$,all whose digits are distinct. The number $N$ does not contain the digit:

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