Question

am i right?!?
What is the sum of the first 19 terms of the series: 63 + 55 + 47 + 39 + 31 + …?

a. -2565

b. -171

c. 247

d. 769.5
My answer: b. -171

Answer 1

\(\Large \color{Black}{\Rightarrow {n \over 2}(a_1 + a_n) }\)

Answer 2

@ParthKohli so was my answer correct??

Answer 3

the formula when you don't have the last term is

\[s _{n} = \frac{n}{2}[2a + n-1)d]\]

a = 1st term, d = common difference and n = number of terms

in your question

a = 63, d = -8 and n = 19

substitute and evaluate

Answer 4

Yeah you're right

Answer 5

First find a19.

\(\Large \color{Black}{\Rightarrow a_{19} = a_1 + -8(19 - 1) }\)
\(\Large \color{Black}{\Rightarrow 63 + -8(18) }\)
\(\Large \color{Black}{\Rightarrow 63 - 144 }\)
\(\Large \color{Black}{\Rightarrow -81 }\)

Answer 6

And there you go correct with the formula

Answer 7

oops n-1 sohould read (n -1)

Answer 8

Yay! so i'm right?!?!

Answer 9

You first need the 19th term, which is

a19 = a1+(n-1)*d

a19 = 63+(19-1)*(-8)

a19 = -81

Then use it in the formula

Sn = (n*(a1+an))/2

S19 = (19*(63+(-81)))/2

S19 = -171

Answer 10

oops, should have put the line S19 = (19*(a1+a19))/2

Answer 11

awesome! thank you!!! :) can you please give @ParthKohli a medal @jim_thompson5910 cuz this only lets me give one out and i want to give both of you one! :D

Answer 12

you're welcome