Arithmetic Progression- help!!

The first term of an arithmetic series is 3 and the common difference is 2. The sum of the first n terms of the series is 360.
Write down an equation satisfied by n. Hence find the value of n.

So a=3 d=2

360=n/2[6+(n-1)2] 720=n[6+(n-1)2] 720=n[6+(2n-2)]
720=6n+2n^2-2
720=2n^2+4n

I don't know what to do next.. Please help! Thanks.

Original post by BubbleLover98

The first term of an arithmetic series is 3 and the common difference is 2. The sum of the first n terms of the series is 360.
Write down an equation satisfied by n. Hence find the value of n.

So a=3 d=2

360=n/2[6+(n-1)2] 720=n[6+(n-1)2] 720=n[6+(2n-2)]
720=6n+2n^2-2
720=2n^2+4n

I don't know what to do next.. Please help! Thanks.

Solve your quadratic equation? You are looking for a positive answer.

(edited 9 years ago)

Reply 2

Jordan97

1

Make this easier for yourself.

When you get to 360=n/2(6+(n-1)2)
Instead of multiplying everything by 2, simply divide the bracket by 2 so you get:
360=n(3+n-1)
Then expand to get
360=3n+n^2-n
Simplify put it equal to 0
n^2+2n-360=0
Now solve the quadratic

Reply 3

TeeEm

19

Original post by BubbleLover98

The first term of an arithmetic series is 3 and the common difference is 2. The sum of the first n terms of the series is 360.
Write down an equation satisfied by n. Hence find the value of n.

So a=3 d=2

360=n/2[6+(n-1)2] 720=n[6+(n-1)2] 720=n[6+(2n-2)]
720=6n+2n^2-2
720=2n^2+4n

I don't know what to do next.. Please help! Thanks.

alternative to Mr M

since n is a positive integer

divide by 2 first and write as n(n+2) = 360

can you think of ... x ... =360 (Whole positive numbers differing by 2)

Reply 4

BubbleLover98

OP

9

so (n-18)(n+20)
So n is 18?

Reply 5

Jordan97

1

Correct!

Arithmetic Progression- help!!

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