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If the sum of the first 3𝑛 positive integers is 150 more than the sum of the first 𝑛 positive integers, then find the sum of the first 4𝑛 positive integers.

 Dec 16, 2021
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The formula for the sum of an arithmetic series is:  Sum  =  n(a + l) / 2

where  n = number of terms     a = first term    l = last term

 

The sum of the first 3n positive integers is:  Sum  =  3n(1 + 3n) / 2

The sum of the first n positive integers is:    Sum  =  n(1 + n) / 2

 

The equation becomes:  3n(1 + 3n) / 2  =  150  +  n(1 + n) / 2

Solving:                               3n(1 + 3n)  =  300 + n(1 + n)

                                               3n + 9n2  =  300 + n + n2

                                              8n2 + 2n - 300  =  0

                                        4n2 + n - 150  =  0

                                     (4n + 25)(n - 6)  =  0

                                                           n  =  6

 

The sum of the first 4n positive integers is:  Sum  =  4n(1 + 4n) / 2

 

Since 4n = 24, the sum is ...

 Dec 17, 2021

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