Question

Can you check where I am wrong in this arithmetic series problem? thanks.

Answer 1

Okay, jus give me a second to type it all out.

Answer 2

\[s _{16} for 12 + 7 + 2 + (-3)\]

Answer 3

First you would find the common difference which is -5

Answer 4

then you need to find the 16th term.. by doing
\[a _{n} = a _{1} (n-1)(d) \]

\[a _{16}= 12(16-1)(-5)\]

Answer 5

and that = -900

Answer 6

now we need to find \[s _{16}\]

Answer 7

\[s _{16} (12 + -900)/2\]

Answer 8

which i get -7104 but thats not right, can you please see where i went wrong.. thanks.

Answer 9

i get that 12 + -900 divide by 2 is -444 then mulitply by 16 and thats how i got -7104

Answer 10

An=A1 +(n-1)d

Answer 11

s16 i believe means the partial sum of the first 16 terms

Answer 12

@Rohitkanna i did do an - a1 + (n-1)(d)... it was 12(16-1)(-5)

Answer 13

\[\sum_{n=1}^{16}12-5(n-1)\]
\[\sum_{n=1}^{16}17-5n\]
\[17\sum_{n=1}^{16}1~-5\sum_{n=1}^{16}n\]
\[17(16)~-5\frac{16(17)}{2}\]

Answer 14

and @amistre64 im not sure what they call s sub 16 but im supposed to find that out by using sn = n(a1 + an)/2

Answer 15

12+(16-2)(-5) is not -900

Answer 16

** (16-1) that is

Answer 17

instead of adding u r multiplying
it shud be 12+(16-1)(-5)
otherwise all ur stepsr correct

Answer 18

15(5) is at best 50+25 ... subtract from 12 is no where near -900

Answer 19

Sn defines the sum of the first n terms of a sequences. the nth partial sum relates to the part of the sequence that is the sum of the first n terms.

Answer 20

@Rohitkannna thanks! i see now what was wrong. It started off wrong.. your right i did multiply instead of adding. I will try it again.

@amistre64 also thanks for also helping.. your right its not -900.. its -63

Answer 21

I get that the answer is -408