1. Find the sum of the first 21 terms of the arithmetic sequence given by the formula an = n+5

I see that I have to plug in numbers and get the terms...but i'm sure there's a shorter way to do this instead of plugging in 21 numbers.

2. The following infinite geometric series will have a finite sum: 1001/6 + 1001/36 + 1001/216 + 1001/1296
Not sure...

http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm

you should know the nth sum of an A.S. is

Sn = n/2 (2a + (n-1)d)
Your sequence has
a=6
d=1

For the G.S. you must have |r| < 1.
What are a,r for this sequence?

Is the answer 336?

Yes,thanks

Pls can u help me to solve: x+y+z=1,x^2+y^2+z^2=34 and x^3+y^3+z^3=94

1. To find the sum of the first 21 terms of the arithmetic sequence given by the formula an = n + 5, you're right that there's a more efficient way than plugging in each number. We need to use the formula for the sum of an arithmetic series.

The sum of an arithmetic series is given by the formula:
Sn = (n/2)(a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.

In this case, the first term a1 is 1 + 5 = 6, and the last term an can be found by substituting n = 21 into the arithmetic sequence formula:
an = 21 + 5 = 26.

So, we have a1 = 6, an = 26, and n = 21. Plugging these values into the formula for the sum of an arithmetic series gives us:
S21 = (21/2)(6 + 26) = (21/2)(32) = 336.

Therefore, the sum of the first 21 terms of the given arithmetic sequence is 336.

2. To determine if the infinite geometric series 1001/6 + 1001/36 + 1001/216 + 1001/1296 will have a finite sum, we need to check if the common ratio between consecutive terms is within the range -1 < r < 1.

The terms of a geometric sequence can be expressed using the formula:
an = a1 * r^(n-1), where an is the nth term, a1 is the first term, r is the common ratio, and n is the term number.

In this case, the first term a1 is 1001/6. To find the common ratio, we can divide any term by its previous term:
(1001/36) / (1001/6) = 1/6.
(1001/216) / (1001/36) = 1/6.
(1001/1296) / (1001/216) = 1/6.

Since the common ratio is indeed 1/6, which satisfies -1 < r < 1, we can conclude that the infinite geometric series will have a finite sum.

To find the sum of the infinite geometric series, we can use the formula:
S = a1 / (1 - r), where S is the sum, a1 is the first term, and r is the common ratio.

Plugging in the values, we get:
S = (1001/6) / (1 - 1/6)
S = (1001/6) / (5/6)
S = (1001/6) * (6/5)
S = 1001/5

Therefore, the sum of the infinite geometric series 1001/6 + 1001/36 + 1001/216 + 1001/1296 is 1001/5.

That's what I get.