Question

AVERAGE OF SERIES/SEQUENCE
The average of 9 consecutive natural numbers is 18.
The highest of these numbers will be
(a) 24
(b) 18 (C) 20 (d) 22
please let me know the steps for solving​

Answer 1

Step-by-step explanation:

The average of 9 consecutive no. is 18

Let the 9 consecutive numbers are x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), (x + 6), (x + 7), and (x + 8).

Average = sum of all no./frequency of the no.

$Average = \frac{x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) + (x + 8)}{9} \\ \implies 18 = \frac{9x + 36}{9} \\ \implies \: 18 \times 9 = 9x + 36 \\ \implies \: 162 - 36 = 9x \\ \implies \: x = { \cancel{\frac{126}{9} }} \\ \therefore \: x = 14$

The highest no. is (x + 8) = 14 + 8 = 22 ans.