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- anonymous

Find the sum of the first 12 terms of the sequence. Show all work for full credit.
1, -4, -9, -14, . . .
please this is my last question and i cant seem to get it right. if possible can someone work it out for me?

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- anonymous

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- campbell_st

what do you know about the question..?

- anonymous

i believe its an arithmetic sequence?

- campbell_st

good start, what's the common difference..?

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- anonymous

ok that where i confuse myself

- campbell_st

well what is the sequence changing by...
common difference = 2nd term - 1st term
check with 3rd term - 2nd term.... is it the same

- anonymous

-5

- campbell_st

great... and the 1st term a = 1
so for the sum of the sequence you can use the formula
\[S_{n} = \frac{n}{2}[2a + (n - 1) \times d]\]
you are given a = 1, you found d = -5 and have been told n = 12
so substitute them and calculate the sum

- anonymous

i get the plugging in but what does d equal? all the rest i plugged in

- campbell_st

d is the common difference d = -5

- anonymous

i see ok 1 min

- anonymous

is it 6/-53

- anonymous

that's what i got but I not entirely sure.

- campbell_st

you are close the calculation becomes
\[S_{12} = \frac{12}{2} \times (-53)\]

- campbell_st

so then
\[S_{12} = 6 \times (-53)\]

- anonymous

so that is the final form?

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