If the numbers from 1 to 45 which are exactly | vedclass

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Question

(D) The numbers from $1$ to $45$ that are exactly divisible by $3$ form an arithmetic progression: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 4

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(D) The numbers from $1$ to $45$ that are exactly divisible by $3$ form an arithmetic progression: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45$.Arranging these in ascending order,the sequence is $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45$.The $1^{st}$ number is $3$,the $2^{nd}$ is $6$,the $3^{rd}$ is $9$,and so on.The $n^{th}$ term is given by $3 	imes n$.For the $9^{th}$ place,the number is $3 	imes 9 = 27$.

If the numbers from $1$ to $45$ which are exactly divisible by $3$ are arranged in ascending order,with the minimum number at the top,which number would be at the ninth place from the top?

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