Question

What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?

605
660
748
816

Answer 1

Is there any one smart enough to help me solve this problem??

Answer 2

\[a _{n} = a _{1} + (n-1)d\]
arithmetic series equation

Answer 3

an = nth term
a1 = first term
d = difference between successive terms
n = which term you are up to

Answer 4

ok

Answer 5

so d = -10
the 12th term = 13 = a12
n = 12
so find a1...@dsolorzano

Answer 6

1 sec

Answer 7

then plug that value into the following equation:
sum = [n (a1 + an) ]/2

Answer 8

how do you solve it?

Answer 9

??

Answer 10

\[a _{n}=a _{1}+(n−1)d, where: a _{12} = 13\]

\[a _{1} = ???\]
\[(n-1)(d) = (12-1)\times (-10)\]

Answer 11

(n-1)(d)=-110

Answer 12

\[a _{12} = a _{1}+(n−1)d\]

Answer 13

\[13 = a _{1}+ (-110)\]

Answer 14

123=a1?

Answer 15

yep.
now sub that into sum = [n (a1 + an) ]/2

Answer 16

123={n(a1+an}/2

Answer 17

\[\sum of series = \frac{ n(a _{1}+a _{n}) }{ 2 }\]

Answer 18

remember 123 is a1 not sum of series

Answer 19

ok

Answer 20

and an = 13
and n =12

Answer 21

12(123+13)/2

Answer 22

yep, equals...?

Answer 23

816

Answer 24

sweet as man

Answer 25

Thanks for your help

Answer 26

so just remember those 2 equations:
an=a1+(n−1)d for the value of term n (in this case the 12th term)
and
sum = n(an+a1)/2

Answer 27

welcome dude