Question

find the sum of first 11 positive number which are multiple of 6​

Answer 1

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Given :-

first 11 positive number which are multiple of 6 are 6 ,12 ,18 , 24 ,30, 36 , 42 , 48 , 54 , 60 , 66

N = 11

D = 6

A = 6

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Answer :-

$sum = \frac{n}{2} (2a + (n - 1)d) \\ \\ = > \frac{11}{2} (2 \times 6 + (11 - 1)6) \\ \\ = > \frac{11}{2} (12 + 60) \\ \\ = > \frac{72 \times 11}{2} \\ \\ = > \frac{792}{2} \\ \\ = > 396$

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Answer 2

Answer:

here is ur answer dear

Step-by-step explanation:

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