# MSC #20: Counting Terms: Arithmetic Sequence

“The first five terms of a sequence are, 2, 9, 16, 23, 30,… What is the 125th term in the sequence? A) 865; B) 870; C) 875; D) 880; E) 885

This question, given in the Junior Level( 13 years old and younger) of the 2^{nd} International Vedic Mathematics Olympiad (IVMO 2022), can be easily solved by using the formula given in textbooks

But many students neither remember the formula nor know what those letters in the formula represent.

If you were asked how many numbers are there from **26** to **89**, would you try to recall the formula taught to us in grade school or would you actually count it?

Now, if you were asked how many numbers there are from **1** to **25**, you **KNOW **that the answer is **25**, because we always start to count from **1** and the last number is the total count.

Since we also **KNOW** that there are **89** numbers from **1** to **89**, removing the first **25** will give us the count from **26** to **89**, which is **89 – 25** **= 64**. In effect we transformed counting from **26** to **89** into counting from **1** to **64**.

I Googled “how to find the number of terms in an arithmetic sequence” and I found this in a website: “**Use the formula**** ****t _{n} = a + (n – 1) d**

**to solve for**

**n****.**Plug in the last term

**(**

**t**_{n}**)**, the first term

**(**

**a****)**, and the common difference

**(**

**d****)**. Work through the equation until you’ve solved for

*.”*

**n**A rearrangement of this formula is easier to use: **n = (t _{n} – a)/d + 1**.

But the key to learning Math is in understanding not memorizing formulas. So, why bother to remember formulas when you can just transform the arithmetic sequence into consecutive counting numbers starting from **1** and immediately know the answer?

In our example, I purposely used “easy-to-compute” numbers so that we can focus on the method not on the calculations: “**Find the number of terms in the sequence 53, 73, 93, …333**.”

It is easy to notice that the common difference between the terms is **20**. So we will first use **by Addition or by Subtraction****(MSC # 15) **to convert the first term into the common difference. ** 53 – 33 = 20**. Thus, we must also deduct **33** from **333** to make it **300.**

We, then divide the “new” first and last terms by the common difference **20**, getting a new sequence of counting numbers from **1** to **15**. Therefore, there are **15** terms in the given sequence.

By reversing the steps in finding the number of terms, we can easily solve the IVMO problem. The common difference is **7. **So first, we must multiply **7** by **125 **to get **875. **But then,to get **2**, the first number of the sequence, we need to deduct** 5 **from** 7. **So, we must also subtract** 5 **from** 875 **to get the answer of **870.**