Question

Find the sum of the first 40 positive integers divisible by 6.
$8 \times 5 \times 2 \div 9 + - 41 + 1 = 0 \times = . \times \div$
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Answer 1

Answer:

a=6,l=40*6=240,n=40 (ARITHMETIC PROGRESSION)

sum of numbers =40/2 × (6+40)=20*46=920

so 920 must be your answer friend

(PLEASE MARK BRAINLIEST)

Answer 2

Step-by-step explanation:

8×5×2÷9+41+1

$= > \: \: 8 \times 5 \times \frac{2}{9} + 41 + 1$

$= > \: \: 40 \times \frac{2}{9} + 41 + 1$

$= > 5 \times 2 + 41 + 1$

$= > 10 + 41 + 1$

$= > 52$