Question

Find three consecutive numbers whose sum is 48.​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf \Rrightarrow Sum \: of \: three \: consecutive \: numbers \: is \: 48.$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf \implies Three \: consecutive \: numbers = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\sf Let \: us \: assume \: that,$

$\sf \Rrightarrow Three \: consecutive \: numbers \: be \: \: (x), \: (x+1), \:( x+2)$

$\sf According \: to \: question,$

$\sf \implies (x)+(x+1)+(x+2)=48$

$\sf \implies 3x+3=48$

$\sf \implies 3x=48 - 3$

$\sf \implies 3x=45$

$\sf \implies x= \cancel \dfrac{45}{3}$

$\sf \implies x= 15$

$\sf \bigstar \: Value \: of \: x = 15$

$\bf \underline {Now},$

$\sf \implies First \: number \: (x) = \bf 15$

$\sf \implies Second \: number \: (x + 1) = \bf (15 + 1) = 16$

$\sf \implies Third \: number \: (x + 1) = \bf (15 + 2) = 17$

$\bf \underline {Therefore},$

$\sf \implies \underline{\boxed{ \sf The \: numbers \: are \: 15, \: 16 \: and \: 17 }}$

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$\bf \underline{ \underline{\maltese\: Verification } }$

To verify the answer just write 15 in place of x.

$\sf \implies (x)+(x+1)+(x+2)=48$

$\sf \implies (15)+(15+1)+(15+2)=48$

$\sf \implies (15)+(16)+(17)=48$

$\sf \implies 15+16+17=48$

$\sf \implies 48=48$

LHS and RHS are equal.

$\bold{\longrightarrow} \:\large{\sf \red{Hence\:Verified\:!}}$