Question

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

Answer 1

To find the sum S4 of the first 4n positive integers, we use the sum formula for an arithmetic series along with the condition that S3 - S1 = 150, solve for 'n', then calculate S4 using the obtained 'n' value.

Step-by-step explanation:

The question involves finding the sum of the first 4n positive integers, given that the sum of the first 3n positive integers is 150 more than the sum of the first n positive integers. To solve this, we use the formula for the sum of an arithmetic series: Sum = n/2(2a + (n - 1)d), where 'n' is the number of terms, 'a' is the first term and 'd' is the common difference. In this case, 'a' equals 1 and 'd' equals 1, as we are dealing with positive integers.

Let's denote S1 as the sum of the first n terms, S3 as the sum of the first 3n terms, and S4 as the sum of the first 4n terms. We know that S3 - S1 = 150, which gives us 3n/2(2 + (3n - 1)) - n/2(2 + (n - 1)) = 150.

By simplifying and solving for n, we find the value that satisfies the equation. Then, using that value of n, we can find the sum S4, which is S4 = 4n/2(2 + (4n - 1)). Applying that value of 'n' will yield the sum of the first 4n positive integers.

Answer 2

Answer: 300

Step-by-step explanation:

`\sum^(3n)_(k=1) k=(3n(3n+1))/(2)\\\\\sum^(n)_(k=1) k=(n(n+1))/(2)\\\\\therefore (3n(3n+1))/(2)-150=(n(n+1))/(2)\\\\3n(3n+1)-300=n(n+1)\\\\9n^2 +3n-300=n^2 +n\\\\8n^2 +2n-300=0\\\\4n^2 +n-150=0\\\\(n-6)(4n+25)=0\\\\n > 0 \implies n=6\\\\\therefore 4n=24\\\\\implies \sum^(4n)_(k=1) k=\sum^(24)_(k=1) k=(24(25))/(2)=300`