arithmetic/geometric sequences, HELP NEEDED BADLY Options

[Gigi]

Yeah. Both my friend and I are totally stuck on this.

3+7+11+15...
How many terms have a sum of less than 500?

Formula for general term (where n=# of terms, a=first term, d=common difference):
Term of n = a+(n-1)d

Okay. So you need to find the general term first, right?
Term of n = 3+(n-1)4
Term of n = 4n-1

Formula for sum of arithmetic sequence (where n=# of terms, a=first term, d=common difference):
Sum of n = n/2(a+term of n)

So that would be...
Sum of n = n/2(3+4n-1)
Sum of n = n/2(4n+2)
Sum of n = 2(n^2)+n

To find the terms with sum less than 500, you'd go:
2(n^2)+n < 500

...right? The thing is. How do you solve that?

--------------------------

Second question. How do you solve for "n" in:

3^(n-1)=729?

[bonluvdan]

for the second question, simply law of indices

3^(n-1) = 729
3^(n-1) = 3^6
n-1=6
n=7

[*kryogenix*]

2n^2+n < 500

n(2n+1) < 500 ?

I guess you'd treat it like solving an inequality.

[Gigi]

Okay, thanks anyway. My friend and I are still clueless and our dumbass of a teacher won't explain anything to us. How she ever became an honours teacher will remain a mystery to me.

[Olive]

I tried the first qn but I did it different to you, hope you have the answers!

Since sum of terms=n/2[2a+(n-1)d]
and n=#of terms, I tried to find n first
ie.
500=n/2 [6+(n-1)4]
1000=n(4n-4+6)
1000=4n(^2)+2n
4n(^2)+2n-1000=0
using quad formula: (which i couldnt be bothered typing with the sqaure root)
n=15.56 and -16.06
since n can only be positive,
n=15.56
where Sn=500
therefore, term sums less than 500, would be nearest to 15.
whew, hope i didnt confuse you further.
;)

[FREEcandies]

She has all the equations (or inequations) right, she just needs to do some algebra to solve for them.

For the inequality (2n^2)+n < 500. You just treat the < sign like a = and solve using the quadratic formula.

For the second problem 3^(n-1)=729, you use the laws of exponents to solve.

n = [729^(1/3)] + 1

[warriors1035]

HOLY SHIT wat grade arre all of u in

im in 7th and this is like hard shit

[Gigi]

This is 10 Honours.

I'm actually in 11 Honours right now, it's just that when my friend asked for my help, I got really into it. Oh, and the 10H teacher refuses to answer any questions, so that could be the problem.

[*mzkandi*]

This is kind old..Gigi will re-open it if she needs to.