Question

find the sum of first 11 positive numbers which are multipliers of 6​

Answer 1

Answer →

Sum of first n natural number is,

$€n = \frac{n(n + 1)}{2}$

Therefore sum of first 11 positive number which are multiple of 6 will be

$= \frac{11(11 + 1)}{2} \times 6 \\ = 396$

Hence, Answer → 396.

Answer 2

So according to question

AP = 6 , 12 .66

a = 6

d = 6

So we know that

Sn = n/2 (2a+(n-1) x d)

S11 = 11/2 (12 + 10 x 6)

S11 = 11/2 x 60

S11 = 330 Ans.

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