Question

3
12. The sum of three consecutive multiples of three is 90. Find the numbers.​

Answer 1

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➠ The sum of three consecutive multiples of three is 90. Find the numbers.

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❏ In this type of question, we have to assume thing whatever is required in such a way that it satisfies all conditions and makes our calculation easy.

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❏ For a number to be multiple of three, number must be divisible by 3 and should follow "3p" where p is an integer.

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❏ Sum of all the multiples of 3 = 90

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Let's Start !

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❖ Here is your answer :)

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☯ Let us assume following thing  

• First multiple of 3 = 3(n - 1) = 3n - 3              { where p = n - 1 }
• Second multiple of 3 = (n - 0) = 3n
• Third multiple of 3 = 3(n + 1) = 3n + 3           { where p = n + 1 }

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As per given condition, sum of these three multiples is 90.

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$:\implies (3n - 3) + 3n + (3n+3) = 90$

$:\implies 3n -3 + 3n + 3n + 3 = 90$

$:\implies 3n + 3n + 3n = 90$

$:\implies 9n = 90$

$:\implies n = \dfrac{90}{9}$

$\boxed{\bold{\orange{:\implies n = 10 }}}$

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As per our assumptions ;

• First multiple of 3 = 3n - 3 = 30 - 3 = 27.      
• Second multiple of 3 = 3n = 30.
• Third multiple of 3  = 3n + 3 = 30 + 3 = 33.

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$\boxed{\boxed{\bold{\therefore Three \; consecutive \; multiples \; are : 27 \; , \; 30 \; , \; 33 }}}$

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Hope this helps u.../

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