Another arithmetic sum series

[radnorgardens]

Question: The first term of an arithmetic series is 16 (a). The 30th term (n) is 100. Calculate S₃₀.

So again, I have the formula:
S_n=n/2[2(a)+(n-1)d].

Using the data above:
S₃₀=30/2[2(16)+(30-1)d]
S₃₀=30/2[32+29d]

Again, I don't understand my next step, or maybe my initial steps are completely wrong. I don't know what to do with the '100', nor 'd' (the common difference). Another attempt:

1st term = 16
30th term = 100

Stuck!

Thanks for any pointers.

 

[Deleted member 4993]

Question: The first term of an arithmetic series is 16 (a). The 30th term (n) is 100. Calculate S₃₀.

So again, I have the formula:
S_n=n/2[2(a)+(n-1)d].

Using the data above:
S₃₀=30/2[2(16)+(30-1)d]
S₃₀=30/2[32+29d]

Again, I don't understand my next step, or maybe my initial steps are completely wrong. I don't know what to do with the '100', nor 'd' (the common difference). Another attempt:

1st term = 16
30th term = 100

Stuck!

Thanks for any pointers.

The equation can be re-written as:

S_n= n/2 * [a₁ + a_n]

you are given a₁ =16 & a₃₀ = 100 → solve for S₃₀

 

[Deleted member 4993]

Something's wrong with that. Check the original problem.
1st term should be 13 OR 100 is the 29th term.
As is, d = 84/29

That is no-doubt a "funky" d - improbable but not impossible.

I believe the objective of the problem was to use the equation I posted - the author probably did not check for value of 'd'.

 

[radnorgardens]

Is the question wrong?

Attached..

 

[stapel]

Attached..

6. The first term of an arithmetic series is 16. The thirtieth term is 100. Calculate S₃₀.

Didn't they give you a formula for the sum, S_n, of the first n terms of a sequence, given the first term a₁ and the n-th term a_n? Why not just plug into that?

 

[radnorgardens]

Thank you, but to understand the formula?

S₃₀ = 30/2 * [a₁ + a_n]
S₃₀ = 15 * [116]
S₃₀ = 1740

Great - thanks, that tallies up with the answer key, but how do you get from:
S_n= n/2 * [2(a)+(n-1)d]
to
S_n= n/2 * [a₁ + a_n] ?

 

[stapel]

....how do you get from:
S_n= n/2 * [2(a)+(n-1)d]
to
S_n= n/2 * [a₁ + a_n] ?

Well, a = a₁, right? And what is the formula for a_n, in terms of a = a₁, n, and d? When you plug that all in to the second formula, what do you get when you simplify down?