Question

What is the sum of the arithmetic sequence 140, 135, 130…, if there are 35 terms?

Answer 1

do you know the formula

Answer 2

isnt this one where you have to use both of the formulas ?

Answer 3

that's better... then tell me both formulas :)

Answer 4

an= a1 + (n-1)d
sn= n/2(a1+an)

Answer 5

like i know you have to use both formulas i just forget hich one you use first

Answer 6

ah that's cool , so we just skipped a big part... now we need to know difference...
do you know how to find it?
if you have
a1=140
a2=135
a3=130

Answer 7

do yo use the first formula first ? the an= a1 + (n-1)d ?

Answer 8

actually I think that we don't need to use first formula at all !
we're dealing only with
\[\LARGE Sn=\frac n2[2a_1+(n-1)d]\]
:)

and we have a1
but we need to find "d"

Answer 9

n= 35 ? right

Answer 10

.. yes last term is a_35
and we only need to find difference , and we're done!

\[\LARGE d=a_n-a_{n-1}\]
\[\LARGE d=a_2-a_1\]
...=?

Answer 11

i got 4934 for d.. i dont think thats right

Answer 12

just use the formula I wrote above..
\[\LARGE d=a_2-a_1\]
\[\LARGE d=135-140\]
\[\LARGE d=?\]

Answer 13

d is -5 ?

Answer 14

yes it is... now substitute back into the formula:
\[\LARGE Sn=\frac n2[2a_1+(n-1)d]\]
\[\LARGE S_{35}=\frac {35}{2}[2\cdot 140+(35-1)\cdot (-5)]\]

Answer 15

1925 ?

Answer 16

That's correct, well done :)

Answer 17

thank you again.. (:

Answer 18

My pleasure :)