Let a1, a2, a3, ... be an arithmetic sequence. Let Sn denote the sum of the first n terms. If S_5 = 1/5 and S_10 = 1/10, then find S_15.
Let's set some variables and make some observations.
Let's let \(d = a_2 -a_1\)
Now, let's note something important. Note that
\(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\)
\(S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\)
\(S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\)
These observations are really important for us to solve the problem.
Let's take two of these cases. We can write a system of equations, and we get
\(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\)
Now, we solve for a1 and d.
Solving this system of equations, we get
\(d = -\dfrac3{250}\\ a_1 = \dfrac8{125}\)
Plugging this into S15, we get
\(S_{15} = 15a_1 + 105d = -\dfrac3{10}\)
So -3/10 is our final answer.
Thanks! :)
Let's set some variables and make some observations.
Let's let \(d = a_2 -a_1\)
Now, let's note something important. Note that
\(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\)
\(S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\)
\(S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\)
These observations are really important for us to solve the problem.
Let's take two of these cases. We can write a system of equations, and we get
\(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\)
Now, we solve for a1 and d.
Solving this system of equations, we get
\(d = -\dfrac3{250}\\ a_1 = \dfrac8{125}\)
Plugging this into S15, we get
\(S_{15} = 15a_1 + 105d = -\dfrac3{10}\)
So -3/10 is our final answer.
Thanks! :)