Question Video: Finding the General Term of a Given Sequence | Nagwa Question Video: Finding the General Term of a Given Sequence | Nagwa

# Question Video: Finding the General Term of a Given Sequence Mathematics • Second Year of Secondary School

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Find, in terms of π, the general term of the sequence (1/4, 9/16, 81/64, 729/256, ...)

03:59

### Video Transcript

Find, in terms of π, the general term of the sequence one-fourth, nine over 16, 81 over 64, 729 over 256, and so on.

In this question, weβre given the first four terms of this sequence. There doesnβt look like thereβs a common difference between the terms, so we could say that this is definitely not an arithmetic sequence. We can check if itβs a geometric sequence which would have a common ratio between the terms by seeing if we can work out what that common ratio would be.

In order to find the ratio π between the first two terms, we would take the second term, nine over 16, and divide by the term before it, one-quarter. We can recall that to divide by one-quarter, that would be equivalent to multiplying by the reciprocal, which would be four over one. We can simplify the four on the numerator and the 16 on the denominator by taking out a factor of four. We then multiply our numerators and denominators. Nine times one would give us nine, and four times one gives us four. We can check if thereβs the same ratio between the third term and the second term. So, we calculate 81 over 64 divided by nine over 16. We can see that the reciprocal of nine over 16 would be 16 over nine. And we simplify the fractions before multiplying the numerators and denominators, which gives us the same common ratio π of nine over four. And if we check this by working out what we need to multiply the third term 81 over 64 by in order to get the fourth term of 729 over 256, we get the same answer of nine over four.

So, letβs think about how we would find the general term of this sequence. We can recall that the general term is another way of saying the πth term. We can recall that the formula to find the πth term π sub π is that π sub π is equal to π times π to the power of π minus one. The π-value represents the common ratio, and the π-value represents the first term of the sequence. Weβve already established that π is equal to nine over four, and the π-value, the first term in the sequence, would be one-quarter. We can then take the values of π and π and plug them into this formula. This gives us π sub π equals one-quarter times nine-quarters to the power of π minus one.

While this is a perfectly valid answer for the πth term of the sequence, it might be worth seeing if we can simplify this right-hand side any further. We can start our simplification by thinking about what happens when we have a power of a fraction. A fraction with a certain exponent value is equivalent to the numerator with that exponent over the denominator with that exponent. If we then consider these fractions multiplied, multiplying the numerators one times nine to the power of π minus one would give us the value on the numerator of nine to the power of π minus one. Multiplying the denominators would give us four times four to the power of π minus one.

If we look at the denominator, we can apply another exponent rule that π₯ to the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus π. On the denominator, the π₯-value here would be four. And our π- and π-values, thatβs the exponents, would be represented by one and π minus one. Adding one and π minus one would leave us with the value of π. So, weβve found a fully simplified answer for the general term of this sequence in terms of π. Itβs nine to the power of π minus one over four to the power of π.

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