Question

the sum of the first n terms in an AP = 3n², how do I find the terms with this information?

Answer 1

AP = arithmetic progression?

first term: \(a_1\)
k-th term: \(a_k = a_1 + (n-1)\Delta\)

sum of the n first terms: \(a_1 + (a_1+\Delta) + (a_1+2\Delta) + \dotsb + (a_1+(n-1)\Delta)\).
So, sum = \(na_1 + (1+2+3+\dotsb+(n-1))\Delta\). And this expression is required to be equal to \(3n^2\).
You need the formula for "(\(1+2+3+\dotsb+n\))".

Answer 2

I have this Sn=3n² and I have to find the sum of first 5 terms and the 10th term, how do I do that?

Answer 3

help?

Answer 4

how do I arrenge the equation?

Answer 5

\(na_1 + \Delta(n(n-1)/2) = 3n^2\). This gives you (an) equation(s).
(for example \(\Delta n^2/2 = 3n^2\) is one of those.
you find the other by comparing the coefficient of \(n\).

Answer 6

I'm still don't get it, but thanks anyway

Answer 7

two polynomials are equal if their coefficients are equal.
\(\Delta/2=3\) -> \(\Delta=6\).
\(na_1 + \Delta (-n)/2 = 0\) (no '\(n\)' term on the RHS)
therefore \(a_1 = \Delta/2 = 3\). now you know everything about this AP.

Answer 8

let me show where I'm stuck

Answer 9

a1=0, an=(n-1), sum=n(n-1)/2, is that right?

Answer 10

Since this is an arithmetic progression, you have to make use of the formula:

1 + 2 + 3 + . . . + (n - 3) + (n - 2) + (n - 1) = n(n - 1)/2

This is merely the Gaussian addition of sequential terms and its derivation should be committed to memory. The problem uses the application of the formula, not its derivation. If you need a quick, non-rigorous derivation, consider the addition of:

1 + 2 + 3 + . . . + (n - 3) + (n - 2) + (n - 1) = S
(n - 1) + (n - 2) + (n - 3) + . . . + 3 + 2 + 1 = S

Notice that each upper/ lower pair is equal to "n" and you have "n - 1" of these, totaling 2S. That is the quick way of grasping the sum. To do an actual proof, you would use mathematical induction. But, again, we are using the formula, not deriving it.

If you consider the equation:\[na _{1} + \frac{ \Delta n(n + 1) }{ 2 } = 3n ^{2}\]this will come down to:\[2na _{1} + \Delta n ^{2} - \Delta n = 6n ^{2}\]For this to make sense, you have to establish values for the coefficients of this equation in the variable "n", shown as a quadratic, but you could simplify it to be linear. Looking at just the "n^2" for the moment, delta has to be "6" because those are the only terms in "n^2". Now that you have that, you can compare:\[2na _{1} = 6n\]to get your\[a _{1}\]

Answer 11

This is merely a little bit more detail, but is exactly the same as the rather fine explanation already put forth by @reemii .

Answer 12

keep in mind that the sum goes from 1 to n-1, so it is Delta(n(n-1))/2 (not n(n+1)/2)

Answer 13

when you set delta=6 we just ignore Δn2−Δn on the equation 2na1+Δn2−Δn=6n2?

Answer 14

sorry, I get it now ¬¬

Answer 15

Yes, that was a typo I had there, but it would be way too long to re-write everything for that one typo. I'll re-write the correction to one small part above:\[na _{1} + \frac{ \Delta n(n - 1) }{ 2 } = 3n ^{2}\]

We didn't "set" delta to 6, we compared the right and left side for the terms in "n^2" and we solved that delta equals 6. Once we solved it, we could then look at the terms in "n".

Answer 16

thanks a lot!

Answer 17

And so,

a1 = 3
a2 = 3 + 6(1)
a3 = 3 + 6(2)

And you're very welcome!