Consider the arithmetic progression: 2+5+8+11+14+···. State 1st term, common difference. Find 50th term. Calculate...

[DidacticBassoon]

Consider the arithmetic progression: 2+5+8+11+14+···
(2)
(i) State the first term and the common difference.
(ii) Calculate the 50th term. (iii) Calculate the sum of the first α terms. (iv) Calculate n where the sum of the first n terms is Sn = 72051

 

[MarkFL]

Hello, and welcome to FMH!

Can you show us what you've got so far, so we know where you're stuck and can best help?

 

[Denis]

Go learn:

Arithmetic Progression

Arithmetic progressions - all formulas.

www.math10.com

 

[Steven G]

Consider the arithmetic progression: 2+5+8+11+14+···
(2)
(i) State the first term and the common difference.
(ii) Calculate the 50th term. (iii) Calculate the sum of the first α terms. (iv) Calculate n where the sum of the first n terms is Sn = 72051

I try not to get upset with students but you want us to tell you the first term. Are you really serious? And you want us to tell you the common difference?
That is you want us to do your homework for you! No, we do not do homework for students. I am sure that you can find a paid service for that.

 

[Steven G]

Go learn:

Arithmetic Progression

Arithmetic progressions - all formulas.

www.math10.com

Denis, let the OP find this link on their own. Maybe you should remove the link.

 

[MarkFL]

Consider the arithmetic progression: 2+5+8+11+14+···
(2)
(i) State the first term and the common difference.
(ii) Calculate the 50th term. (iii) Calculate the sum of the first α terms. (iv) Calculate n where the sum of the first n terms is Sn = 72051

Since several days have gone by with surprisingly no further feedback from the OP, I am going to follow up as I like to do when time permits to give the thread some utility for people who've found it via a search engine.

i) The first term is 2, and the common difference is 3. This should be obvious from the given terms.

ii) The \(n\)th term is \(a_n=3n-1\), and so \(a_{50}=149\).

iii) [MATH]S_{\alpha}=\sum_{k=1}^{\alpha}(3k-1)=3\frac{\alpha(\alpha+1)}{2}-\alpha=\frac{\alpha(3\alpha+1)}{2}[/MATH]
iv) [MATH]S_n=72051[/MATH]
[MATH]\frac{n(3n+1)}{2}=72051[/MATH]
[MATH]n(3n+1)=144102[/MATH]
[MATH]3n^2+n-144102=0[/MATH]
[MATH](n-219)(3n+658)=0[/MATH]
Discard the negative root:

[MATH]n=219[/MATH]