Question

the average of the first nine integral multiples of 3 is

Answer 1

$the \: first \: nine \: integral \\ \: multiplies \: of \: 3 \: are \\ 3.6.9.12.15.18.21.24.27 \\ so \: their \: average - \\ \frac{total \: of \: quantities}{no. \: of \: quantities} \\ = \frac{3 + 6 + ... + 27}{9} \\ = \frac{ \frac{9}{2} (3 + 27)}{9} \\ = 15$

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Answer 2

the first nine integral multiplies of 3are 3.6.9.12.15.18.21.24.27 so their average−

no.ofquantities/

totalofquantities

=

9

3+6+...+27

=

9

2

9

(3+27)

=15