Sequences + Series.... PLEASE HELP?

Question

any help please! and if you could be so kind as to show work or provide explanations...I WILL ETERNALLY LOVE YOU!

1.) what is the tenth term and the sum of the first ten terms of:
a.) 10, 7, 4, 1, ...
b.) 4, -2, 1, 0.5, ...

2.) what is the infinite sum of: 5, 5/3, 5/9, 5/27, ...?

thank you so much. please help!

Answer 1

1a) The nth term is [10 - 3*(n-1)], so the 10th term is -17. The sum of the first 10 terms is 10*[average value of terms]
= 10*[ (first + last)/2 ] = 10 * [-7 / 2] = -35

1b) Doesn't make sense unless 2nd term is positive or 4th term is negative.

2) Infinite geometric series. Assume sum exists and is equal to S. So:
S = 5 + 5/3 + 5/9 + 5/27 + ...
S/3 = 5/3 + 5/9 + 5/27 + ...

Note that the fractional terms in the above series are identical, so subtracting leaves just the 5:
S - S/3 = 5

Now solve for S:
S - S/3 = (2/3)S = 5
S = 5*3/2 = 15/2 [or 7.5 if you prefer]

Answer 2

In any arithmetic sequence we've 1st term a 2d term a + d third term a + 2d 4th term a + 3-d ..... ... nth term a + (n-a million)d (make certain that's sparkling on your head) we are advised that the 1st term is 4 so a=4 all of us know from above that the 4th term is a + 3-d additionally we are able to make certain that the nineteenth term is a + 18d now the nineteenth term is 4 situations the 4th term so a + 18d = 4(a +3-d) a + 18d = 4a +12d 6d = 3a 6d = 3(4) (a = 4) 6d = 12 d = 2 now the thirtieth term is a + 29d that's 4 + 29(2) = 4 + fifty 8 =sixty two and your carried out

Answer 3

Regarding #2) First, a quote from my textbook:

A geometric series ∑ a*r^(n-1) ......n=1 to infinity

= a + ar + a*r^2 + a*r^3 + ... is convergent if l r l < 1, and its

sum is a / (1-r).
-------------------------------------
For #2, the series is 5 + 5*(1/3) + 5*(1/3)^2 + 5*(1/3)^3 ...

This is a geometric series with a =5 and common ratio r = (1/3)

The sum is equal to 5 / [ 1 - (1/3) ] = 5 / (2/3) = 15/2 = 7.5