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Given that the sum of all the multiples of 3 from 3 to 99 is t and the
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07 Feb 2022, 04:00
07 Feb 2022, 04:00
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Given that the sum of all the multiples of 3 from 3 to 99 inclusive is t, and the sum of the multiples of 6 from 6 to 198 inclusive is s, what is s in terms of t?
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2t
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Given that the sum of all the multiples of 3 from 3 to 99 is t and the
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24 Feb 2022, 22:33
24 Feb 2022, 22:33
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Carcass wrote:
Given that the sum of all the multiples of 3 from 3 to 99 inclusive is t, and the sum of the multiples of 6 from 6 to 198 inclusive is s, what is s in terms of t?
Show: ::
2t
Sum of terms in AP = \(\frac{n}{2}[2(a) + (n - 1)d]\)
Number of terms in multiples = \(\frac{(L - F)}{multiple} + 1\)
Number of multiples of 3 = \(\frac{(99 - 3)}{3} + 1 = 33\)
Number of multiples of 6 = \(\frac{(198 - 6)}{6} + 1 = 33\)
Sum of multiples of 3 = \(\frac{33}{2}[2(3) + (33 - 1)3] = \frac{33}{2}[6 + 96] = 1,683\)
Sum of multiples of 6 = \(\frac{33}{2}[2(6) + (33 - 1)6] = \frac{33}{2}[12 + 192] = 3,366\)
Clearly, Sum of multiples of 6 is twice the sum of multiples of 3
Therefore, s = 2t
Re: Given that the sum of all the multiples of 3 from 3 to 99 is t and the
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07 Mar 2024, 06:52
07 Mar 2024, 06:52
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