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# Algebra

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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.

Jun 14, 2024

#1
+1250
+1

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! :)

Jun 14, 2024

#1
+1250
+1

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! :)

NotThatSmart Jun 14, 2024