Find the sum of first 8 multiples of 3.
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3+6+9+12+15+18+21+24
due to the divisibility of 3 ,it will be divisible by 3(3,6,9)
3(1+2+3+4+5+6+7+8)
3×36
108
due to the divisibility of 3 ,it will be divisible by 3(3,6,9)
3(1+2+3+4+5+6+7+8)
3×36
108
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Answered by
3
first 8 multiples of 3 are
3,6,9,12,15,18,21,24
their sum= 3+6+9+12+15+18+21+24=108
or
3×(8×9)÷2
3×(72÷2)
3×36
=108
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3,6,9,12,15,18,21,24
their sum= 3+6+9+12+15+18+21+24=108
or
3×(8×9)÷2
3×(72÷2)
3×36
=108
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