Question

the sum of three consecutive multiples of 7 is 63 find these multilples​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 14,21 \: and \: 28}}}$

Step-by-step explanation:

Let the three consecutive multiples of 7 be x, (x + 7) and (x + 14) respectively.

According to the question ;

x + x + 7 + x + 14 = 63

⇒ 3x + 21 = 63

⇒ 3x = 63 - 21

⇒ 3x = 42

⇒ x = $\sf \dfrac{42}{3}$

⇒ x = 14

Thus, the consecutive multiples of 7 are;

• x = 14
• x + 7 = 14 + 7 = 21
• x + 14 = 14 + 14 = 28

Hence, the required three consecutive multiples of 7 is 14, 21 and 28.

Answer 2

$\Huge{\boxed{Solution}}$

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Given :

Sum of three consecutive

multiple of 7 is 63

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To find :

Multiples

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Let the :

Multiples be 'x'. So

three consecutive

multiples of 7 will

be x , (x + 7) and  (x + 14)

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According to the question

we have to solve this

equation x+ x+7 + x+14

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Lets do it,

⇒ x + x + 7 + x + 14 = 63

⇒ 3x + 21 = 63

⇒ 3x = 63 - 21

⇒ 3x = 42

⇒ x = $\dfrac{42}{3}$

⇒ x = 14

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Verification :

⇒ 14 + 14 + 7 + 14 + 14 = 63

⇒ 28 + 7 + 14 + 14 = 63

⇒ 28 + 7 + 28 = 63

⇒ 56 + 7  =  63

⇒  63 = 63

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Here, L.H.S = R.H.S

Hence it is verified

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First consecutive multiple

= x

= 14

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Second consecutive multiple

= x + 7

= 14 + 7

= 21

=======================

Third consecutive multiple

= x + 14

= 14 + 14

= 28

=======================

Therefore three consecutive multiple

of 7 is 14, 21 and 28.