Question

How do you find the $9^{\mathrm{th}}$ term of the arithmetic sequence $a_{10}=100$ and $a_{20}=50$?

Answer 1

Finding the $9^{\mathrm{th}}$ term of an arithmetic sequence:

Step 1: Assumption and formation of two linear equations.

Let the first term and the common difference of the series be $a$ and $d$ respectively.

We know that for an arithmetic sequence with the first term $a$ and common difference $d$, the ${\mathrm{n}}^{\mathrm{th}}$ term will be $a_n=a+\left(n-1\right)d$.

So, the ${10}^{\mathrm{th}}$ term will be:

$$\begin{gathered} a_{10}=a+\left(10-1\right)d \\ =a+9d \end{gathered}$$

Now, by the question, $a_{10}=100$.

Thus, we get: $a+9d=100$ $...\left(1\right)$

Again, the ${20}^{\mathrm{th}}$ term will be,

$$\begin{gathered} a_{20}=a+\left(20-1\right)d \\ =a+19d \end{gathered}$$

Now, by the question, $a_{20}=50$.

Thus, we get: $a+19d=50$ $...\left(2\right)$

Step 2: Solving these linear equations

Subtracting the equation $\left(1\right)$ form the equation $\left(2\right)$, we get:

$$\begin{gathered} a+19d-a-9d=50-100 \\ \Rightarrow 10d=-50 \\ \Rightarrow d=-\frac{50}{10} \\ \therefore d=-5 \end{gathered}$$

Then, from equation $\left(1\right)$, we get:

$$\begin{gathered} a=100-9d \\ =100-9\times \left(-5\right) \\ =100+45 \\ =145 \end{gathered}$$

Therefore, the first term of the series is $a=145$ and the common difference is $d=-5$.

Step 3: Finding the $9^{\mathrm{th}}$ term

Using the formula, $a_n=a+\left(n-1\right)d$, we get the $9$th term:

$$\begin{gathered} a_9=a+\left(9-1\right)d \\ =145+8\times \left(-5\right) \\ =145-40 \\ =105 \end{gathered}$$

Therefore, the $9^{\mathrm{th}}$ term of the given arithmetic series is $a_9=105$.