Question

Help , I don’t know how to solve this

Answer 1

Answers in bold:

S9 = 2

i = 20

R = -2

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Step-by-step explanation:

`S_0 = 20` is the initial term because your teacher mentioned
`A_0 = I` as the initial term.

Then R = -2 is the common difference because we subtract 2 from each term to get the next term. In other words, we add -2 to each term to get the next term.

Here is the scratch work for computing terms S1 through S4.

`\begin{array}\cline{1-2}S_(n) = S_(n-1) - 2 & S_(n) = S_(n-1) - 2\\S_(1) = S_(1-1) - 2 & S_(2) = S_(2-1) - 2\\S_(1) = S_(0) - 2 & S_(2) = S_(1) - 2\\S_(1) = 20 - 2 & S_(2) = 18 - 2\\S_(1) = 18 & S_(2) = 16\\\cline{1-2}S_(n) = S_(n-1) - 2 & S_(n) = S_(n-1) - 2\\S_(3) = S_(3-1) - 2 & S_(4) = S_(4-1) - 2\\S_(3) = S_(2) - 2 & S_(4) = S_(3) - 2\\S_(3) = 16 - 2 & S_(4) = 14 - 2\\S_(3) = 14 & S_(4) = 12\\\cline{1-2}\end{array}`

Then here is S5 though S8

`\begin{array}l\cline{1-2}S_(n) = S_(n-1) - 2 & S_(n) = S_(n-1) - 2\\S_(5) = S_(5-1) - 2 & S_(6) = S_(6-1) - 2\\S_(5) = S_(4) - 2 & S_(6) = S_(5) - 2\\S_(5) = 12 - 2 & S_(6) = 10 - 2\\S_(5) = 10 & S_(6) = 8\\\cline{1-2}S_(n) = S_(n-1) - 2 & S_(n) = S_(n-1) - 2\\S_(7) = S_(7-1) - 2 & S_(8) = S_(8-1) - 2\\S_(7) = S_(6) - 2 & S_(8) = S_(7) - 2\\S_(7) = 8 - 2 & S_(8) = 6 - 2\\S_(7) = 6 & S_(8) = 4\\\cline{1-2}\end{array}`

And finally we arrive at S9.

`S_(n) = S_(n-1) - 2\\\\S_(9) = S_(9-1) - 2\\\\S_(9) = S_(8) - 2\\\\S_(9) = 4 - 2\\\\S_(9) = 2\\\\`

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Because we have an arithmetic sequence, there is a shortcut.

`a_n` represents the nth term

S9 refers to the 10th term because we started at index 0. So we plug n = 10 into the arithmetic sequence formula below.

`a_n = a_1 + d(n-1)\\\\a_n = 20 + (-2)(n-1)\\\\a_n = 20 - 2(n-1)\\\\a_(10) = 20 - 2(10-1)\\\\a_(10) = 20 - 2(9)\\\\a_(10) = 20 - 18\\\\a_(10) = 2\\\\`

In other words, we start with 20 and subtract off 9 copies of 2 to arrive at 20-2*9 = 20-18 = 2, which helps see a faster way why
`S_9 = 2`