# Thread: Stuck on progressions question

1. ## Stuck on progressions question

How many terms of the series 4,10,16, ....... must be taken such that the sum is equal to 602?

So far i have:

Sn=n/2(2a+(n-1)d)
602x2=n(2x4+(n-1)6)
1204=n(8+6n-6)
1204=8n+6n squared-6n
1204=6n squared+2n
602=3n squared+n
n+3n squared -602=0

I have got my self into a bit of a pickle with this now and am stuck any help would be much appreciated

cheers

2. $\displaystyle 3n^{2}+n-602 = 0$

You can solve the equation by factorization

Notice $\displaystyle -602 X 3 = -1806 = 43X-42$

$\displaystyle 3n^{2}+43n-42n-602 = 0$

Obtain the roots of this equation thus, you'll get the answer.

If you're having trouble factorizing, you could use the formula for the roots of a quadratic equation $\displaystyle ax^{2}+bx+c = 0$

$\displaystyle x=(-b-(b^{2}-4ac)^{1/2})/2a$ and $\displaystyle x=(-b+(b^{2}-4ac)^{1/2})/2a$

3. Hello, boboulas!

You're work is correct . . .

How many terms of the series 4,10,16 . . .must be taken
. . such that the sum is equal to 602?

So far i have:

$\displaystyle S_n\:=\:\frac{n}{2}[2a+(n-1)d]$

$\displaystyle 602\cdot2\:=\:n[2\cdot4+(n-1)6]$

$\displaystyle 1204\:=\:n(8+6n-6)$

$\displaystyle 1204\:=\:8n+6n^2-6n$

$\displaystyle 1204\:=\:6n^2+2n$

$\displaystyle 602\:=\:3n^2 +n$

$\displaystyle n+3n^2-602\:=\:0$

I have got myself into a bit of a pickle . . .

You have a quadratic equation: .$\displaystyle 3n^2 + n - 602 \:=\:0$

Factor (or use the Quadratic Formula): .$\displaystyle (n-14)(3n+43) \:=\:0$

. . And we have: .$\displaystyle n \:=\:14,\:-\frac{43}{3}$

Since $\displaystyle \,n$ must be a positive integer: .$\displaystyle n = 14$