Question

Math question! Please help! 100 points

An arithmetic sequence has first term a = 5 and common difference d = 2. How many terms of this sequence must be added to get 1440?

Answer 1

General formula

• a_n=a+(n-1)d
• s_n=n/2[2a+(n-1)d

We need n

• n/2[2(5)+2(n-1)]=1440
• n[10+2n-2]=2880
• n[2n+8]=2880
• 2n²+8n-2880=0
• n²+4n-1440=0
• (n+40)(n-36)=0

Take it positive

• n=36

Answer 2

To find how many terms of this sequence must be added to get 1440, we must know the arithmetic sequence equation to find any term in the sequence:

⇒
`a_(n)=a_(1) +(n-1)d`

• 
  `a_(n)` --> value of the nth number of the sequence
• 
  `a_(1)` --> first term of the sequence
• n --> position of the nth term
• d --> common difference

Let's examine the information given:

• first term a = 5

⇒
`a_(1) =5`

• common difference d = 2

⇒ d = 2

Therefore the equation for finding the nth term of the sequence so far is:

`a_(n)=5+(n-1)*2=5+2n-2=3+2n`

Now we want to find how many terms this sequence must be added to get 1440

General equation for adding all the terms =
`(n(a_(1)+a_(n) ) )/(2)`

• 
  `a_(1)`: first term of the sequence
• 
  `a_(n)`: last term of the sequence

• n: number of terms in the sequence.

Using all the information given, let's plug in the all the values:

`1440=(n*(5+(3+2n)))/(2) \\1440=(n*(8+2n))/(2) \\2880=8n+2n^2\\2n^2+8n-2880=0\\(2n+80)(n-36)=0`

To solve this we set (2n + 80) = 0 and (n-36) = 0

`2n+80=0\\ 2n = -80\\n= -40`
`n - 36 =0\\n = 36`

In this case, n can only equal 36 since n cannot be negative.

Answer: 36 terms of this sequence must be added to get 1440.

Hope that helps!