Question

Find three consecutive odd integers such that the sum of the first number, two less than the second number and three more than the third number is 70

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• Three consecutive odd integers.

• Sum of the first number, two less than the second number and three more than the third number is 70

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$\red{\underline \bold{To \: Find:}}$

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• The consecutive odd integers.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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Let us observe some consecutive odd integers.

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• 1 , 3 , 5

• -7 , -5 , -3

• 9 , 11 , 13

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Here we can clearly observe some similarities among the groups.

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$\underline{\bold{\texttt{Above groups can be rewritten as:}}}$

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• 1 , '1 + 2' , '1 + 4'

• -7 , '-7 + 2' , '-7 + 4'

• 9 , '9 + 2' , '9 + 4'

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Let 'x' be the first number then second number will be 'x + 2' and third number will be 'x + 4'

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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$\purple\longrightarrow$ $\sf x + x + x + 7 = 70$

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$\sf \longmapsto 3x + 7 = 70$

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$\sf \longmapsto 3x = 63$

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$\sf \longmapsto x = \dfrac { 63 } { 3 }$

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$\bf \dashrightarrow x = 21$

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$\underline{\bold{\texttt{Consecutive odd integers will be:}}}$

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• 21

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• 21 + 2

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• 21 + 4

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Hence three consecutive odd integers will be 21 , 23 , 25.

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$\rule{200}5$