# Sequences III

**“A sequence starts, 4, 5, 8, 13, 20,… What is the 98th term in this sequence?**

**A) 9511 B) 9413 C) 9697 D) 9707 E) 9805”**

This question was given in the Intermediate category of the 3^{rd} International Vedic Mathematics Olympiad (IVMO 2023) and very few participants failed to get the correct answer.

In our previous post, “ Sequence II (https://www.math-inic.com/blog/sequence-ii/), we discussed this question from the Primary group of the 1^{st} International Vedic Mathematics Olympiad (IVMO 2021):

**“What are the next two numbers in this sequence?**

**97 79 64 52 __ ___**

**A 42 36 B 44 38 C 44 37 D 43 38 E 43 37”**

This is also a quadratic sequence but is easier to solve because we are only required to find the **next two** numbers in the sequence but in the present case, we are asked to give the 98^{th} term. So, we really need to study the techniques of solving quadratic sequences.

The given sequence is : 4 5 8 13 20

The first differences are: 1 3 5 7

The second differences are: 2 2 2

The common second difference of 2 between the first differences confirm that the sequence is quadratic in the form of An^2 + Bn + C and **A = 1**. (We will explain why A is half the common second difference in our next post about sequences)

The first differences are increasing by 2, so the difference between the **1 ^{st} term** and the

**0**of the sequence must also be 2 or 1 – 2 = -1. So, the 0

^{th}term^{th}term must be [4 – (-1) = 5). Using this value in A(0)^2 + B(0) + C = 5, we can see that C = 5.

Our quadratic expression for the terms in the sequence now become n^2 + Bn + 5.

For n = 1, we have 1^2 + B(1) + 5 = 4, B = -2

To check, for n= 2, 2^2 -2(1) + 5 = 5

For n = 3, 3^2 + -2(3) + 5 = 8

Now to get the 98^{th} term, we just have to solve (98)^2 -2(98) + 5.

98^2 is easily obtained by using the Vedic Math sub-sutra, “** Whatever the deficiency, reduce it further by that amount and set-up the square of the deficiency**”.

The deficiency of 98 from the base, 100, is 2. Thus, we have 98 – 2 = 96 for the first part of the square and 2^2 or 04 for the second part. Note that we must allow to decimal places for the second part. 98^2 = 9604

(98)^2 -2(98) + 5 = 9604 – 196 + 5

= 9413

We can also solve this using base multiplication:

(98)^2 -2(98) + 5 = 98(98 – 2) + 5

= 98 (96) + 5

= (98 – 4 |02 x 04) + 5

= 9408 + 5

= 9413.

Choice B is the correct answer.