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Find the sum of the sequence227,225,223,221,...,209

User Amram
by
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1 Answer

7 votes

2180 (option A)

Step-by-step explanation:

common difference = next term - previous term

d = 225 - 227

d = -2

The formula for sum of sequence of an arithmetic progression:


S_n=(n)/(2)\lbrack2a_1+(n-1)d\rbrack
\begin{gathered} \\ a_n=a_1+(n-1)d \\ S_n=(n)/(2)(a_1+a_n) \\ a_n\text{ = 209, }a_{1\text{ }}=\text{ 227} \end{gathered}
\begin{gathered} S_n\text{ =}(n)/(2)(209\text{ + 227)} \\ S_n\text{ =}(n)/(2)(436\text{)} \end{gathered}
\begin{gathered} To\text{ get n, we would apply the formula:} \\ a_n=a_1\text{ + (n-1)d} \\ a_n\text{ = last term = 209} \\ n\text{ = ?, d = -2, }a_1=\text{ 227} \\ 209\text{ = 227 + (n - 1)(-2)} \\ \end{gathered}
\begin{gathered} 209\text{ = 227 -2n + 2} \\ 209\text{ - 227 = -2n + 2} \\ -18\text{ = -2n + 2} \\ -18-2\text{ = -2n} \\ -20\text{ = -2n} \\ n\text{ = -20/-2} \\ n\text{ = 10} \end{gathered}
\begin{gathered} \text{The sum of the sequence = S}_n\text{ = }(n)/(2)(436) \\ S_n=(10)/(2)*436 \\ S_n\text{ = 2180 (option A)} \end{gathered}

User Benjamin Baumann
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