Question

the sum of three consecutive multiples of 7 is 63 find the multiples

Answer 1

7 14 21 can be the multiples of 7

Answer 2

Hey!!!

$let \: the \: first \: consecutive \: no. \: be \: x + 7$
$therefore \: second \: no \: = \: x + 14$
$and \: the \: last \: no \: = \: x + 21$
$as \: per \: condition$
$(x + 7) + (x + 14) + (x + 21) = 63$
$x + 7 + x + 14 + x + 21 = 63$
$3x + 42 = 63$
$3x = 21$
$x = 7$
Therefore,

$the \: first \: no \: is \: 7 + 7 = 14$
$second \: no \: is \: 7 + 14 = 21$
$third \: no \: is \: 7 + 21 = 28$


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