Question

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

Answer 1

2180 (option A)

Explanation:

common difference = next term - previous term

d = 225 - 227

d = -2

The formula for sum of sequence of an arithmetic progression:

$S_n=\frac{n}{2}\lbrack2a_1+(n-1)d\rbrack$$\begin{gathered} \\ a_n=a_1+(n-1)d \\ S_n=\frac{n}{2}(a_1+a_n) \\ a_n\text{ = 209, }a_{1\text{ }}=\text{ 227} \end{gathered}$$\begin{gathered} S_n\text{ =}\frac{n}{2}(209\text{ + 227)} \\ S_n\text{ =}\frac{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{ = }\frac{n}{2}(436) \\ S_n=\frac{10}{2}\times436 \\ S_n\text{ = 2180 (option A)} \end{gathered}$