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Sequences III

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 3rd 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 1st 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 98th 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 1st term and the 0th term  of the sequence must also be 2  or 1 – 2 = -1. So, the 0th 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 98th 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.

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