Question Video: Finding the Terms of a Sequence given Its General Term and the Value of a Term Mathematics

Find the first five terms of the sequence π‘Ž_𝑛, given π‘Ž_(𝑛 + 1) = ((βˆ’1)^𝑛)/(9π‘Ž_𝑛), 𝑛 β‰₯ 1 and π‘Žβ‚ = βˆ’11.

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Video Transcript

Find the first five terms of the sequence π‘Ž sub 𝑛, given π‘Ž sub 𝑛 plus one is equal to negative one to the 𝑛th power divided by nine multiplied by π‘Ž sub 𝑛, where 𝑛 is greater than or equal to one and π‘Ž sub one equals negative 11.

In this question, we are given the value of the first term of our sequence π‘Ž sub one, which is equal to negative 11. We are asked to find the first five terms. So we therefore need to calculate the second, third, fourth, and fifth terms of the sequence. These are denoted π‘Ž sub two, π‘Ž sub three, π‘Ž sub four, and π‘Ž sub five.

The formula we are given is an example of a recursive formula, where π‘Ž sub 𝑛 plus one is given in terms of π‘Ž sub 𝑛. We can therefore calculate any term of the sequence by substituting in the previous term. The second term π‘Ž sub two is therefore equal to negative one to the power of one divided by nine multiplied by π‘Ž sub one. We know that π‘Ž sub one is equal to negative 11. This expression is equal to negative one over negative 99, which in turn simplifies to one over 99. This is the second term of the sequence, and we can use this to calculate the third.

Using our recursive formula once again, π‘Ž sub three is equal to negative one squared over nine multiplied by π‘Ž sub two. We know that π‘Ž sub two is one over 99. On the denominator, we have nine multiplied by one over 99, which is the same as nine over 99. Dividing the numerator and denominator by nine, this simplifies to one over 11 or one eleventh. As negative one squared is equal to one, π‘Ž sub three is equal to one divided by one eleventh. We know that dividing by a fraction is the same as multiplying by the reciprocal of this fraction. One divided by one eleventh is the same as one multiplied by 11 over one. This is equal to 11. π‘Ž sub three, the third term in our sequence, is 11.

Next, we need to calculate the fourth term π‘Ž sub four. This is equal to negative one cubed divided by nine multiplied by π‘Ž sub three. And we have just calculated that π‘Ž sub three equals 11. The fourth term of the sequence is therefore equal to negative one over 99, recalling that cubing a negative number gives a negative answer. Finally, we have π‘Ž sub five is equal to negative one to the fourth power divided by nine multiplied by π‘Ž sub four. We can substitute negative one over 99 for π‘Ž sub four. And this in turn simplifies to one over negative one eleventh. Using a similar method that we used to calculate π‘Ž sub three, we have π‘Ž sub five is equal to negative 11. The first five terms of the sequence are negative 11, one over 99, 11, negative one over 99, and negative 11.

We notice that the first term is equal to the fifth term. This means that we have a periodic sequence where the first four terms are repeated. π‘Ž sub one is equal to π‘Ž sub five, which is equal to π‘Ž sub nine and so on. Likewise, the second, sixth, and 10th terms are equal. The same is true for the third, seventh, and 11th terms, together with the fourth, eighth, and 12th terms.

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