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The second and sixth terms of a geometric progression are 3/10 and 243/160, respectively. What is the first term of this sequence?

Question

The second and sixth terms of a geometric progression are 310 \frac{3}{10} and 243160 \frac{243}{160} , respectively. What is the first term of this sequence?

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Solution

In a geometric progression, the ratio between any two consecutive terms is constant. This ratio is often referred to as the common ratio.

Given that the second term is 3/10 and the sixth term is 243/160, we can find the common ratio by taking the sixth term divided by the second term and then taking the fourth root of the result (because the sixth term is four terms away from the second term).

So, the common ratio (r) is ((243/160) / (3/10))^(1/4) = 3/2.

The first term (a) can be found by dividing the second term by the common ratio. So, a = (3/10) / (3/2) = 2/10 = 1/5.

Therefore, the first term of this geometric progression is 1/5.

This problem has been solved

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